Exercise Questions
1. A man has to go 50m due north, 40m due east and 20m due south to reach a field.
(a)What distance he has to walk to reach the field?
(b)What is his displacement from his house to the field?
Solution:
Step 1: Identify the Path Segments
The motion consists of three distinct parts. To analyze these, we treat the movements as individual segments of a total path.
- $d_1 = 50\text{ m}$ (North)
- $d_2 = 40\text{ m}$ (East)
- $d_3 = 20\text{ m}$ (South)
Step 2: Calculate Total Distance
The sources define distance as a scalar quantity representing the total area an object covers in motion. It is calculated by summing all segments using the formula $d=d_1+d_2+d_3$.
- $d = 50\text{ m} + 40\text{ m} + 20\text{ m}$
- Total Distance = $\mathbf{110\text{ m}}$
Step 3: Establish a Coordinate System for Displacement
Displacement is a vector quantity that indicates how far an object is from its destination. According to the sources, it is expressed as $x = x_f – x_i$, where $x_f$ is the final position and $x_i$ is the initial position. We set the starting point (house) as the origin $(0,0)$.
- After moving North: $(0, 50)$
- After moving East: $(40, 50)$
- After moving South: $(40, 50 – 20) = (40, 30)$
The final position $x_f$ is $(40, 30)$.
Step 4: Calculate the Magnitude of Displacement
Using the coordinates of the initial position $(0,0)$ and final position $(40,30)$, we find the magnitude (the straight-line distance):
- $|x| = \sqrt{(40 – 0)^2 + (30 – 0)^2}$
- $|x| = \sqrt{1600 + 900} = \sqrt{2500}$
- Magnitude = $\mathbf{50\text{ m}}$
Step 5: Determine the Direction
The sources emphasize that for displacement, it is mandatory to specify the direction of travel. We find the angle $\theta$ relative to the East axis:
- $\tan \theta = \frac{\text{Vertical Displacement}}{\text{Horizontal Displacement}} = \frac{30}{40} = 0.75$
- $\theta = \tan^{-1}(0.75) \approx \mathbf{37^\circ}$
- Final Displacement: $\mathbf{50\text{ m}}$ at $\mathbf{37^\circ \text{ North of East}}$.
2. A particle starts from the origin, goes along the X-axis to the point \( (20\,\text{m}, 0) \) and then returns along the same line to the point \( (-20\,\text{m}, 0) \). Find the distance and displacement of the particle during the trip.
Solution:
Step 1: Description of Motion
- Initial position: \( (0,0) \)
- Motion from origin to +20 m: \( 0 \rightarrow +20\,\text{m} \)
- Return motion from +20 m to −20 m: \( +20 \rightarrow -20\,\text{m} \)
(a) Distance Travelled
Distance is the total length of the actual path
travelled by the particle.
\[
\text{Distance} = 20 + 40 = \boxed{60\,\text{m}}
\]
(b) Displacement
Displacement is the change in position of the particle
from start to end.
\[
\text{Displacement} = x_f – x_i = (-20) – 0 = \boxed{-20\,\text{m}}
\]
The negative sign indicates that the displacement is along the
negative X-axis.
Distance by road = \(320 \, \text{km}\)
Time taken by aeroplane = \(30 \, \text{minutes} = 0.5 \, \text{hour}\)
Time taken by bus = \(8 \, \text{hours}\)
\[
\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}}
\]
\[
= \frac{260}{0.5} = 520 \, \text{km/h}
\]
\text{Average speed} = \frac{320}{8} = 40 \, \text{km/h}
\]
\[
\text{Average velocity} = \frac{\text{Displacement}}{\text{Time}}
\]
\[
= \frac{260}{0.5} = 520 \, \text{km/h}
\]
Direction: from Patna to Ranchi
\[
\text{Average velocity} = \frac{260}{8} = 32.5 \, \text{km/h}
\]
Final meter reading = \(12416 \, \text{km}\)
Total time taken = \(2 \, \text{hours}\)
\text{Distance} = 12416 – 12352 = 64 \, \text{km}
\]
\[
\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}}
\]
\[
= \frac{64}{2} = 32 \, \text{km/h}
\]
Therefore, displacement \(= 0\).
\[
\text{Average velocity} = \frac{\text{Displacement}}{\text{Time}} = \frac{0}{2} = 0 \, \text{km/h}
\]
\[
u = 0
\]
\[
v = 18 \times \frac{5}{18} = 5 \, \text{m/s}
\]
\[
t = 2 \, \text{s}
\]
\[
v = u + at
\]
5 = 0 + a(2)
\]
a = \frac{5}{2} = 2.5 \, \text{m/s}^2
\]
From the graph:
Initial speed, \(u = 0 \, \text{m/s}\)
Final speed, \(v = 20 \, \text{m/s}\)
Time taken, \(t = 8 \, \text{s}\)
Distance travelled is equal to the area under the speed–time graph.
The graph forms a triangle.
\text{Distance} = \frac{1}{2} \times \text{base} \times \text{height}
\]
\[
= \frac{1}{2} \times 8 \times 20 = 80 \, \text{m}
\]
\[
a = \frac{v – u}{t}
\]
\[
= \frac{20 – 0}{8} = 2.5 \, \text{m/s}^2
\]
Acceleration of the car = \(2.5 \, \text{m/s}^2\)
\[
u = 0
\]
\[
a = 5 \, \text{m/s}^2
\]
\[
v = u + at
\]
v = 0 + 5(10)
\]
v = 50 \, \text{m/s}
\]
a = 0 \Rightarrow v = 50 \, \text{m/s (constant)}
\]
a = -5 \, \text{m/s}^2
\]
v = 50 + (-5)(10)
\]
v = 0
\]
\text{Distance} = \frac{1}{2} \times (30 + 10) \times 50
\]
= \frac{1}{2} \times 40 \times 50
\]
= 1000 \, \text{m}
\]
(a) The acceleration
(b) The distance travelled in 0 to 10 s
(c) The displacement in 0 to 10 s
\[
a = \frac{v – u}{t}
\]
\[
= \frac{8 – 2}{10}
= 0.6 \, \text{m/s}^2
\]
\[
\text{Area} = \frac{1}{2} \times ( \text{sum of parallel sides} ) \times \text{height}
\]
\[
= \frac{1}{2} \times (2 + 8) \times 10
= 50 \, \text{m}
\]
displacement = area under the graph.
\[
\text{Displacement} = 50 \, \text{m}
\]
9. The given figure shows the graph of x-coordinate of a particle going along the X-axis as a function of time. Find:
(a) The average velocity during 0 to 10 s
(b) Instantaneous velocity at 2 s, 5 s, 8 s and 12 s.
Displacement from 0 to 10 s:
Initial position at \(t = 0\), \(x = 0\)
Final position at \(t = 10\), \(x = 100 \, \text{m}\)
\text{Average velocity} = \frac{\text{Displacement}}{\text{Time}}
\]
\[
= \frac{100}{10} = 10 \, \text{m/s}
\]
\[
\text{Slope} = \frac{50 – 0}{2.5 – 0}
= \frac{50}{2.5}
= 20 \, \text{m/s}
\]
At \(t = 5 \, \text{s}\) (horizontal segment):
Slope \(= 0\)
\[
v = 0 \, \text{m/s}
\]
\[
\text{Slope} = \frac{100 – 50}{10 – 7.5}
= \frac{50}{2.5}
= 20 \, \text{m/s}
\]
\[
\text{Slope} = \frac{0 – 100}{15 – 10}
= \frac{-100}{5}
= -20 \, \text{m/s}
\]
\text{Distance} =
\left( \frac{1}{2} \times 20 \times 5 \right)
+
\left( \frac{1}{2} \times 20 \times 5 \right)
\]
= 50 + 50 = 100 \, \text{m}
\]
\text{Displacement} =
\left( \frac{1}{2} \times 20 \times 5 \right)
+
\left( -\frac{1}{2} \times 20 \times 5 \right)
\]
= 50 – 50 = 0
\]
V_{\text{avg}} = \frac{\text{Displacement}}{\text{Time}}
\]
V_{\text{avg}} = \frac{0}{40} = 0 \, \text{m/s}
\]
11. The given figure shows x–t graph of a particle. Find the time \(t\) such that the average velocity of the particle during the period 0 to \(t\) is zero.
Solution:
Average velocity is given by:
\[
\text{Average velocity} = \frac{\text{Displacement}}{\text{Time}}
\]
For average velocity to be zero, displacement must be zero.
From the graph:
At \(t = 0\), \(x = 20\).
The particle again reaches \(x = 20\) at \(t = 12 \, \text{s}\).
Hence, displacement from 0 to 12 s is zero.
Required time, \(t = 12 \, \text{s}\).
12. A particle starts from a point A and travels along the solid curve shown in the given figure. Find approximately the position B of the particle such that the average velocity between the positions A and B has the same direction as the instantaneous velocity at B.
Solution: Average velocity between A and B is along the straight line joining A and B.
Instantaneous velocity at B is along the tangent to the curve at point B.
For both velocities to have the same direction, the straight line AB must be parallel to the tangent at B.
Hence, point B is approximately point (5,3) on the curve where the tangent drawn at B is parallel to the line joining A and B.
13. An object having a velocity 4.0 m/s is accelerated at the rate of 1.2 m/s² for 5.0 s. Find the distance travelled during the period of acceleration.
Solution:
Given:
\[
u = 4 \, \text{m/s}, \quad a = 1.2 \, \text{m/s}^2, \quad t = 5 \, \text{s}
\]
Distance travelled is given by:
\[
s = ut + \frac{1}{2}at^2
\]
\[
s = (4)(5) + \frac{1}{2}(1.2)(5)^2
\]
\[
s = 20 + \frac{1}{2}(1.2)(25)
\]
\[
s = 20 + 15 = 35 \, \text{m}
\]
Distance travelled = \(35 \, \text{m}\).
14. A person travelling at 43.2 km/h applies the brake giving a deceleration of 6.0 m/s² to his scooter. How far will it travel before stopping?
Solution:
Convert speed into m/s:
\[
u = 43.2 \times \frac{5}{18} = 12 \, \text{m/s}
\]
Final velocity,
\[
v = 0
\]
Acceleration (retardation),
\[
a = -6 \, \text{m/s}^2
\]
Using the equation:
\[
v^2 = u^2 + 2as
\]
\[
0^2 = (12)^2 + 2(-6)s
\]
\[
0 = 144 – 12s
\]
\[
12s = 144
\]
\[
s = 12 \, \text{m}
\]
Distance travelled before stopping = \(12 \, \text{m}\).
15. A train starts from rest and moved with a constant acceleration of 2.0 m/s² for half a minute. The brakes are then applied and the train comes to rest in one minute.
(a) Find the total distance moved by the train
(b) The maximum speed attained by the train
(c) The position(s) of the train at half the maximum speed.
Solution:
(a) Total Distance Moved
The motion is divided into two parts: acceleration and deceleration.
Part 1: Acceleration (\(S_1\))
- Initial velocity (\(u\)) = \(0\)
- Acceleration (\(a\)) = \(2 \text{ m/s}^2\)
- Time (\(t\)) = \(30 \text{ sec}\)
- \(S_1 = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(2)(30)^2 = 900 \text{ m}\)
Part 2: Deceleration (\(S_2\))
- First, find the maximum velocity:
\(v = u + at = 0 + (2)(30) = 60 \text{ m/s}\) - For braking phase:
\(u’ = 60 \text{ m/s}, \quad v’ = 0, \quad t’ = 60 \text{ sec}\) - Deceleration:
\(a’ = \frac{v’ – u’}{t’} = \frac{0 – 60}{60} = -1 \text{ m/s}^2\) - \(S_2 = \frac{v’^2 – u’^2}{2a’} = \frac{0 – 60^2}{2(-1)} = 1800 \text{ m}\)
Total Distance:
\(S = S_1 + S_2 = 900 + 1800 = 2700 \text{ m} = 2.7 \text{ km}\)
(b) Maximum Speed Attained
The maximum speed is reached just before braking:
\(v_{\text{max}} = 60 \text{ m/s}\)
(c) Positions at Half the Maximum Speed
Half of the maximum speed = \(30 \text{ m/s}\). This occurs twice:
- During Acceleration:
\(v^2 – u^2 = 2aS\)
\(30^2 = 2(2)S \Rightarrow 900 = 4S \Rightarrow S = 225 \text{ m}\) - During Deceleration:
\(S = \frac{v^2 – u^2}{2a} = \frac{30^2 – 60^2}{2(-1)} = 1350 \text{ m}\)
Total position from start:
\(900 + 1350 = 2250 \text{ m} = 2.25 \text{ km}\)
16. A bullet travelling with a velocity of 16 m/s penetrates a tree trunk and comes to rest in 0.4 m. Find the time taken during the retardation.
Solution:
Given:
\[
u = 16 \, \text{m/s}, \quad v = 0, \quad s = 0.4 \, \text{m}
\]
Using the equation of motion:
\[
v^2 = u^2 + 2as
\]
\[
0^2 = (16)^2 + 2(a)(0.4)
\]
\[
0 = 256 + 0.8a
\]
\[
a = -320 \, \text{m/s}^2
\]
Using:
\[
v = u + at
\]
\[
0 = 16 + (-320)t
\]
\[
t = \frac{16}{320} = 0.05 \, \text{s}
\]
Time taken during retardation = \(0.05 \, \text{s}\).
17. A bullet going with speed 350 m/s enters a concrete wall and penetrates a distance 5.0 cm before coming to rest. Find the deceleration.
Solution:
Given:
\[
u = 350 \, \text{m/s}, \quad v = 0, \quad s = 5 \times 10^{-2} \, \text{m}
\]
Using the equation of motion:
\[
v^2 = u^2 + 2as
\]
\[
0^2 = (350)^2 + 2(a)(0.05)
\]
\[
0 = 122500 + 0.1a
\]
\[
a = -1.225 \times 10^6 \, \text{m/s}^2
\]
Deceleration of the bullet = \(1.225 \times 10^6 \, \text{m/s}^2\).
18. A particle starting from rest moves with constant acceleration. If it takes 5.0 s to reach the speed 18.0 km/h. Find
(a) The average velocity during this period
(b) The distance travelled by the particle during this period.
Solution:
Given:
\[
u = 0, \quad t = 5 \, \text{s}
\]
Final velocity:
\[
v = 18 \times \frac{5}{18} = 5 \, \text{m/s}
\]
Using:
\[
v = u + at
\]
\[
5 = 0 + a(5)
\]
\[
a = 1 \, \text{m/s}^2
\]
(b) Distance travelled
\[
s = ut + \frac{1}{2}at^2
\]
\[
s = 0 + \frac{1}{2}(1)(5)^2
\]
\[
s = 12.5 \, \text{m}
\]
(a) Average velocity
\[
V_{\text{avg}} = \frac{\text{Distance}}{\text{Time}}
\]
\[
V_{\text{avg}} = \frac{12.5}{5} = 2.5 \, \text{m/s}
\]
Average velocity = \(2.5 \, \text{m/s}\),
Distance travelled = \(12.5 \, \text{m}\).
19. A driver takes 0.20 s to apply the brakes after he sees a need for it. This is called the reaction time of the driver. If he is driving a car at a speed of 54 km/h and the brakes cause a deceleration of 6.0 m/s², find the distance travelled by the car after he sees the need to put the brakes on.
Solution:
Speed of the car:
\[
v = 54 \times \frac{5}{18} = 15 \, \text{m/s}
\]
Distance travelled during reaction time:
\[
s_1 = vt = 15 \times 0.2 = 3 \, \text{m}
\]
When brakes are applied:
\[
u = 15 \, \text{m/s}, \quad v = 0, \quad a = -6 \, \text{m/s}^2
\]
Using the equation:
\[
v^2 = u^2 + 2as
\]
\[
0^2 = (15)^2 + 2(-6)s_2
\]
\[
0 = 225 – 12s_2
\]
\[
s_2 = 18.75 \, \text{m}
\]
Total distance travelled:
\[
s = s_1 + s_2 = 3 + 18.75 = 21.75 \, \text{m}
\]
\[
\approx 22 \, \text{m}
\]
Total distance travelled after seeing the need to brake = \(22 \, \text{m}\).
20. Complete the following table:
| Car Model | Driver X | Driver Y |
|---|---|---|
| A Deceleration on hard braking = 6.0 m/s² |
Speed = 54 km/h Braking distance a = ……………………. Total stopping distance b = …………………….. |
Speed = 54 km/h Braking distance a = ……………………. Total stopping distance b = …………………….. |
| Speed = 72 km/h Braking distance c = ……………………. Total stopping distance d = …………………….. |
Speed = 72 km/h Braking distance c = ……………………. Total stopping distance d = …………………….. |
|
| B Deceleration on hard braking = 7.5 m/s² |
Speed = 54 km/h Braking distance e = ……………………. Total stopping distance f = …………………….. |
Speed = 54 km/h Braking distance e = ……………………. Total stopping distance f = …………………….. |
| Speed = 72 km/h Braking distance g = ……………………. Total stopping distance h = …………………….. |
Speed = 72 km/h Braking distance g = ……………………. Total stopping distance h = …………………….. |
| Car Model | Driver X (Reaction time = 0.20 s) | Driver Y (Reaction time = 0.30 s) |
|---|---|---|
| A Deceleration on hard braking = 6.0 m/s² |
Speed = 54 km/h Braking distance a = 18.75 m Total stopping distance b = 21.75 m |
Speed = 54 km/h Braking distance a = 18.75 m Total stopping distance b = 23.25 m |
| Speed = 72 km/h Braking distance c = 33.33 m Total stopping distance d = 37.33 m |
Speed = 72 km/h Braking distance c = 33.33 m Total stopping distance d = 39.33 m |
|
| B Deceleration on hard braking = 7.5 m/s² |
Speed = 54 km/h Braking distance e = 15.0 m Total stopping distance f = 18.0 m |
Speed = 54 km/h Braking distance e = 15.0 m Total stopping distance f = 19.5 m |
| Speed = 72 km/h Braking distance g = 26.67 m Total stopping distance h = 30.67 m |
Speed = 72 km/h Braking distance g = 26.67 m Total stopping distance h = 32.67 m |
Explanation:
To complete the table, we calculate three quantities:
1. Speed in m/s
2. Braking distance
3. Reaction distance and total stopping distance
Step 1: Convert speed from km/h to m/s
\[
u = \text{Speed in km/h} \times \frac{5}{18}
\]
For 54 km/h:
\[
u = 54 \times \frac{5}{18} = 15 \, \text{m/s}
\]
For 72 km/h:
\[
u = 72 \times \frac{5}{18} = 20 \, \text{m/s}
\]
Step 2: Formula for braking distance
Braking distance is given by:
\[
s_b = \frac{u^2}{2a}
\]
where
\(u\) = initial speed,
\(a\) = deceleration on hard braking.
Car Model A (deceleration = 6.0 m/s²)
For speed 54 km/h (\(u = 15 \, \text{m/s}\)):
\[
s_b = \frac{15^2}{2 \times 6}
= \frac{225}{12}
= 18.75 \, \text{m}
\]
For speed 72 km/h (\(u = 20 \, \text{m/s}\)):
\[
s_b = \frac{20^2}{2 \times 6}
= \frac{400}{12}
= 33.33 \, \text{m}
\]
Car Model B (deceleration = 7.5 m/s²)
For speed 54 km/h (\(u = 15 \, \text{m/s}\)):
\[
s_b = \frac{15^2}{2 \times 7.5}
= \frac{225}{15}
= 15.0 \, \text{m}
\]
For speed 72 km/h (\(u = 20 \, \text{m/s}\)):
\[
s_b = \frac{20^2}{2 \times 7.5}
= \frac{400}{15}
= 26.67 \, \text{m}
\]
Step 3: Reaction distance
Reaction distance is given by:
\[
s_r = u \times t_r
\]
where \(t_r\) is reaction time.
Driver X (reaction time = 0.20 s)
For 54 km/h:
\[
s_r = 15 \times 0.20 = 3.0 \, \text{m}
\]
For 72 km/h:
\[
s_r = 20 \times 0.20 = 4.0 \, \text{m}
\]
Driver Y (reaction time = 0.30 s)
For 54 km/h:
\[
s_r = 15 \times 0.30 = 4.5 \, \text{m}
\]
For 72 km/h:
\[
s_r = 20 \times 0.30 = 6.0 \, \text{m}
\]
Step 4: Total stopping distance
\[
\text{Total stopping distance} = s_b + s_r
\]
Examples:
Car A, Driver X, 54 km/h:
\[
= 18.75 + 3.0 = 21.75 \, \text{m}
\]
Car A, Driver Y, 72 km/h:
\[
= 33.33 + 6.0 = 39.33 \, \text{m}
\]
Car B, Driver X, 72 km/h:
\[
= 26.67 + 4.0 = 30.67 \, \text{m}
\]
Car B, Driver Y, 54 km/h:
\[
= 15.0 + 4.5 = 19.5 \, \text{m}
\]
Conclusion:
Higher speed and longer reaction time increase the stopping distance,
while higher deceleration (better braking system) reduces the stopping distance.
21. A police jeep is chasing a culprit going on a motorbike. The motorbike crosses a turning at a speed of 72 km/h. The jeep follows it at a speed of 90 km/h, crossing the turning ten seconds later than the bike. Assuming that they travel at constant speeds, how far from the turning will the jeep catch up with the bike?
Solution:
Speed of the motorbike:
\[
V_{\text{bike}} = 72 \times \frac{5}{18} = 20 \,\text{m/s}
\]
Speed of the police jeep:
\[
V_{\text{jeep}} = 90 \times \frac{5}{18} = 25 \,\text{m/s}
\]
Distance travelled by the motorbike in 10 seconds:
\[
\text{Distance} = \text{Speed} \times \text{Time} = 20 \times 10 = 200 \,\text{m}
\]
Relative speed of the jeep with respect to the bike:
\[
V_{\text{relative}} = 25 – 20 = 5 \,\text{m/s}
\]
Time taken by the jeep to catch the bike:
\[
t = \frac{\text{Relative distance}}{\text{Relative speed}} = \frac{200}{5} = 40 \,\text{s}
\]
Distance travelled by the jeep in this time:
\[
\text{Distance} = 25 \times 40 = 1000 \,\text{m} = 1 \,\text{km}
\]
Answer: The police jeep will catch the motorbike at a distance of 1 km from the turning.
\[
V_1 = 60 \times \frac{5}{18} = 16.6 \text{ m/s}
\]
\[
V_2 = 42 \times \frac{5}{18} = 11.6 \text{ m/s}
\]
\[
V_{\text{rel}} = 16.6 – 11.6 = 5 \text{ m/s}
\]
\[
d_{\text{rel}} = 5 + 5 = 10 \text{ m}
\]
\[
t = \frac{d_{\text{rel}}}{V_{\text{rel}}} = \frac{10}{5} = 2 \text{ s}
\]
\[
d = 16.6 \times 2 = 33.2 \text{ m}
\]
\[
= 33.2 + 5 = 38.2 \text{ m} \approx 38 \text{ m}
\]
23. A ball is projected vertically upward with a speed of \(50 \text{ m/s}\). Find:
(a) the maximum height reached,
(b) the time taken to reach the maximum height, and
(c) the speed at half the maximum height.
(Take \(g = 10 \text{ m/s}^2\))
Solution
Given:
- Initial velocity (\(u\)) = \(50 \text{ m/s}\)
- Acceleration due to gravity (\(g\)) = \(-10 \text{ m/s}^2\)
- Final velocity at highest point (\(v\)) = \(0\)
(a) Maximum Height Reached (\(H\))
Using the third equation of motion:
\(v^2 – u^2 = 2aH\)
\(0 – (50)^2 = 2(-10)H\)
\(-2500 = -20H\)
\(H = \frac{2500}{20} = 125 \text{ m}\)
(b) Time taken to reach maximum height (\(t\))
Using the first equation of motion:
\(v = u + at\)
\(0 = 50 + (-10)t\)
\(10t = 50\)
\(t = 5 \text{ sec}\)
(c) Speed at half the maximum height
Half of the maximum height:
\(s’ = \frac{125}{2} = 62.5 \text{ m}\)
Using the third equation of motion:
\(v’^2 – u^2 = 2as’\)
\(v’^2 – (50)^2 = 2(-10)(62.5)\)
\(v’^2 – 2500 = -1250\)
\(v’^2 = 1250\)
\(v’ = \sqrt{1250} = 35.35 \text{ m/s} \approx 35 \text{ m/s}\)
24. A ball is dropped from a balloon going up at a speed of \(7 \text{ m/s}\). If the balloon was at a height of \(60 \text{ m}\) at the time of dropping the ball, how long will the ball take to reach the ground?
(Take \(g = 10 \text{ m/s}^2\))
Solution:
Important Concept: When the ball is dropped, it inherits the velocity of the balloon. Since the balloon is moving upward, the initial velocity of the ball is also upward.
Given:
- Initial velocity (\(u\)) = \(-7 \text{ m/s}\) (upward negative, downward positive)
- Displacement (\(s\)) = \(60 \text{ m}\) (downward)
- Acceleration (\(a\)) = \(10 \text{ m/s}^2\)
Equation of Motion:
\(s = ut + \frac{1}{2}at^2\)
\(60 = -7t + \frac{1}{2}(10)t^2\)
\(60 = -7t + 5t^2\)
Rearranging into quadratic form:
\(5t^2 – 7t – 60 = 0\)
Using quadratic formula:
\(t = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}\)
\(t = \frac{-(-7) \pm \sqrt{(-7)^2 – 4(5)(-60)}}{2(5)}\)
\(t = \frac{7 \pm \sqrt{49 + 1200}}{10}\)
\(t = \frac{7 \pm \sqrt{1249}}{10}\)
\(t \approx \frac{7 \pm 35.34}{10}\)
Final Result:
- Taking positive root:
\(t = \frac{7 + 35.34}{10} = 4.2 \text{ sec}\)
The ball takes approximately \(4.2 \text{ seconds}\) to reach the ground.
25. A stone is thrown vertically upward with a speed of \(28 \text{ m/s}\).
(a) Find the maximum height reached by the ball.
(b) Find its velocity one second before it reaches the maximum height.
(c) Does the answer to part (b) change if the initial speed is more than \(28 \text{ m/s}\) (e.g., \(40 \text{ m/s}\) or \(80 \text{ m/s}\))?
Solution
Given:
- Initial velocity (\(u\)) = \(28 \text{ m/s}\)
- Final velocity at maximum height (\(v\)) = \(0\)
- Acceleration due to gravity (\(g\)) = \(-9.8 \text{ m/s}^2\)
(a) Maximum Height Reached (\(S\))
Using the equation:
\(v^2 – u^2 = 2aS\)
\(0 – (28)^2 = 2(-9.8)S\)
\(-784 = -19.6S\)
\(S = \frac{784}{19.6} = 40 \text{ m}\)
(b) Velocity one second before reaching maximum height
First, find the time to reach maximum height:
\(v = u + at\)
\(0 = 28 – 9.8t\)
\(t = \frac{28}{9.8} \approx 2.85 \text{ sec}\)
Time one second before highest point:
\(t’ = t – 1 = 1.85 \text{ sec}\)
Velocity at this time:
\(v’ = u + at’\)
\(v’ = 28 – (9.8)(1.85)\)
\(v’ = 28 – 18.13 = 9.87 \text{ m/s}\)
(c) Does the answer change for higher initial speed?
No.
The velocity one second before reaching the maximum height will always be \(9.8 \text{ m/s}\), regardless of the initial speed.
This is because during the last one second of upward motion, the velocity decreases uniformly under gravity:
\(v = v’ + at \Rightarrow 0 = v’ – (9.8)(1)\)
\(v’ = 9.8 \text{ m/s}\)
26. A person sitting on the top of a tall building is dropping balls at regular intervals of one second. Find the positions of the 3rd, 4th, and 5th balls when the 6th ball is being dropped.
(Take \(g = 9.8 \text{ m/s}^2\))
Solution
Key Concept: When the 6th ball is being dropped (\(t = 0\)), the earlier balls have already been falling for different times:
- 5th ball: falling for \(1 \text{ s}\)
- 4th ball: falling for \(2 \text{ s}\)
- 3rd ball: falling for \(3 \text{ s}\)
Given:
- Initial velocity (\(u\)) = \(0\)
- Acceleration (\(a\)) = \(9.8 \text{ m/s}^2\)
- Distance formula: \(S = \frac{1}{2}gt^2\)
1. Position of the 3rd Ball (\(t = 3 \text{ s}\))
\(S_3 = \frac{1}{2}(9.8)(3)^2 = \frac{1}{2}(9.8)(9) = 44.1 \text{ m}\)
The 3rd ball is \(44.1 \text{ m}\) below the top.
2. Position of the 4th Ball (\(t = 2 \text{ s}\))
\(S_4 = \frac{1}{2}(9.8)(2)^2 = \frac{1}{2}(9.8)(4) = 19.6 \text{ m}\)
The 4th ball is \(19.6 \text{ m}\) below the top.
3. Position of the 5th Ball (\(t = 1 \text{ s}\))
\(S_5 = \frac{1}{2}(9.8)(1)^2 = 4.9 \text{ m}\)
The 5th ball is \(4.9 \text{ m}\) below the top.
Final Answer:
- 3rd Ball: \(44.1 \text{ m}\) below top
- 4th Ball: \(19.6 \text{ m}\) below top
- 5th Ball: \(4.9 \text{ m}\) below top
27. A healthy young man standing at a distance of \(7 \text{ m}\) from an \(11.8 \text{ m}\) high building sees a kid slipping from the top floor. With what speed (assumed uniform) should he run to catch the kid at the arms height \(1.8 \text{ m}\)?
Solution
The man must reach the building in the same time it takes the kid to fall to the arms height.
1. Time taken by the kid to fall
- Initial velocity (\(u\)) = \(0 \text{ m/s}\)
- Initial height = \(11.8 \text{ m}\)
- Final height = \(1.8 \text{ m}\)
Displacement:
\(s = 11.8 – 1.8 = 10 \text{ m}\)
- Acceleration (\(g\)) = \(9.8 \text{ m/s}^2\)
Using equation of motion:
\(s = ut + \frac{1}{2}gt^2\)
\(10 = 0 + \frac{1}{2}(9.8)t^2\)
\(10 = 4.9t^2\)
\(t^2 = \frac{10}{4.9} \approx 2.04\)
\(t \approx 1.43 \text{ s}\)
2. Speed of the man
- Distance = \(7 \text{ m}\)
- Time = \(1.43 \text{ s}\)
Using:
\(v = \frac{\text{Distance}}{\text{Time}}\)
\(v = \frac{7}{1.43} \approx 4.9 \text{ m/s}\)
Final Answer: The man should run with a uniform speed of \(4.9 \text{ m/s}\).
28. An NCC parade is going at a uniform speed of 6 km/h through a place under a berry tree on which a bird is sitting at a height of 12.1 m. At a particular instant the bird drops a berry. Which cadet (give the distance from the tree at the instant) will receive the berry on his uniform?
(Take \(g = 9.8 \text{ m/s}^2\))
Solution
To catch the mango, the cadet must reach the point directly below the tree at the same time the mango reaches the ground.
1. Time taken by the mango to fall
- Initial velocity (\(u\)) = \(0\)
- Height (\(s\)) = \(19.6 \text{ m}\)
- Acceleration (\(g\)) = \(9.8 \text{ m/s}^2\)
Using:
\(s = ut + \frac{1}{2}gt^2\)
\(19.6 = 0 + \frac{1}{2}(9.8)t^2\)
\(19.6 = 4.9t^2\)
\(t^2 = 4 \Rightarrow t = 2 \text{ s}\)
2. Distance covered by the cadet
- Speed = \(6 \text{ km/h} = \frac{5}{3} \text{ m/s}\)
- Time = \(2 \text{ s}\)
Using:
\(d = vt\)
\(d = \frac{5}{3} \times 2 = \frac{10}{3} \approx 3.33 \text{ m}\)
Final Answer: The cadet should be at a distance of \(3.33 \text{ m}\) from the tree.
29. A ball is dropped from a height. If it takes \(0.2 \text{ s}\) to cross the last \(6 \text{ m}\) before hitting the ground, find the total height from which it was dropped.
(Take \(g = 9.8 \text{ m/s}^2\))
Solution:
1. Analysis of the last 6 m
- Distance (\(S\)) = \(6 \text{ m}\)
- Time (\(t\)) = \(0.2 \text{ s}\)
- Acceleration (\(a\)) = \(9.8 \text{ m/s}^2\)
- Initial velocity (\(u\)) = ?
Using:
\(S = ut + \frac{1}{2}at^2\)
\(6 = u(0.2) + \frac{1}{2}(9.8)(0.2)^2\)
\(6 = 0.2u + 4.9(0.04)\)
\(6 = 0.2u + 0.196\)
\(u = \frac{6 – 0.196}{0.2} = \frac{5.804}{0.2} = 29 \text{ m/s}\)
2. Motion before the last 6 m
- Initial velocity = \(0 \text{ m/s}\)
- Final velocity = \(29 \text{ m/s}\)
- Acceleration = \(9.8 \text{ m/s}^2\)
Using:
\(v^2 – u^2 = 2aS\)
\(29^2 = 2(9.8)x\)
\(841 = 19.6x\)
\(x = \frac{841}{19.6} \approx 42.05 \text{ m}\)
3. Total Height
\(\text{Total Height} = x + 6 = 42.05 + 6 = 48.05 \text{ m}\)
Final Answer: The total height is approximately \(48 \text{ m}\).
30. A ball is dropped from a height of \(5 \text{ m}\) onto a sandy floor and penetrates the sand up to \(10 \text{ cm}\) before coming to rest. Find the retardation of the ball in the sand.
(Take \(g = 9.8 \text{ m/s}^2\))
Solution
The motion is divided into two parts: free fall in air and motion in sand.
1. Motion in Air (Free Fall)
- Initial velocity (\(u\)) = \(0\)
- Distance (\(s\)) = \(5 \text{ m}\)
- Acceleration (\(a\)) = \(9.8 \text{ m/s}^2\)
Using:
\(v^2 = u^2 + 2as\)
\(v^2 = 0 + 2(9.8)(5) = 98\)
\(v = \sqrt{98} \approx 9.89 \text{ m/s}\)
Velocity just before entering sand = \(9.89 \text{ m/s}\)
2. Motion in Sand (Retardation)
- Initial velocity (\(u_1\)) = \(9.89 \text{ m/s}\)
- Final velocity (\(v_1\)) = \(0\)
- Distance (\(s\)) = \(0.1 \text{ m}\)
Using:
\(v_1^2 – u_1^2 = 2as\)
\(0 – (9.89)^2 = 2(a)(0.1)\)
\(-98 = 0.2a\)
\(a = -490 \text{ m/s}^2\)
Final Answer: The retardation of the ball in sand is \(490 \text{ m/s}^2\).
31. An elevator is descending with uniform acceleration. To measure the acceleration, a person in the elevator drops a coin at the moment the elevator starts. The coin is 6 ft above the floor of the elevator at the time it is dropped. The person observes that the coin strikes the floor in 1 second. Calculate from these data the acceleration of the elevator.
(Take \(g = 9.8 \text{ m/s}^2\))
Solution:
1. Motion of the coin
- Initial velocity (\(u\)) = \(0\)
- Time (\(t\)) = \(1 \text{ s}\)
- Acceleration = \(9.8 \text{ m/s}^2\)
Using:
\(S_c = ut + \frac{1}{2}gt^2\)
\(S_c = 0 + \frac{1}{2}(9.8)(1)^2 = 4.9 \text{ m}\)
2. Motion of the elevator floor
- Initial velocity (\(u\)) = \(0\)
- Time (\(t\)) = \(1 \text{ s}\)
- Acceleration = \(a\)
Using:
\(S_e = ut + \frac{1}{2}at^2\)
\(S_e = \frac{1}{2}a(1)^2 = \frac{a}{2}\)
3. Condition for collision
The coin meets the floor when:
\(S_c = 1.8 + S_e\)
\(4.9 = 1.8 + \frac{a}{2}\)
4. Solving for \(a\)
\(4.9 – 1.8 = \frac{a}{2}\)
\(3.1 = \frac{a}{2}\)
\(a = 6.2 \text{ m/s}^2\)
Conversion to ft/s\(^2\):
\(a = 6.2 \times 3.28 \approx 20.34 \text{ ft/s}^2\)
Final Answer: The acceleration of the elevator is \(6.2 \text{ m/s}^2\) (approximately \(20.34 \text{ ft/s}^2\)).
32. A ball is thrown horizontally from a point 100 m above the ground with a speed of 20 m/s. Find (a) the time it takes to reach the ground, (b) the horizontal distance it travels before reaching the ground, (c) the velocity (direction and magnitude) with which it strikes the ground.
(Take \(g = 9.8 \text{ m/s}^2\))
Solution:
1. Time taken to reach the ground
For horizontal projection, time depends only on height:
\(t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 100}{9.8}} = 4.51 \text{ s}\)
2. Horizontal range
Horizontal velocity remains constant:
\(x = ut = 20 \times 4.51 \approx 90 \text{ m}\)
3. Velocity at impact
- Horizontal component (\(V_x\)) = \(20 \text{ m/s}\)
- Vertical component (\(V_y\)):
\(V_y = gt = 9.8 \times 4.51 = 44.1 \text{ m/s}\)
Resultant velocity:
\(V = \sqrt{V_x^2 + V_y^2} = \sqrt{20^2 + (44.1)^2} \approx 48.42 \text{ m/s}\)
Direction:
\(\tan \beta = \frac{V_y}{V_x} = \frac{44.1}{20} = 2.205\)
\(\beta = \tan^{-1}(2.205) \approx 66^\circ\)
Final Answer: The ball strikes the ground with a velocity of \(48.42 \text{ m/s}\) at an angle of \(66^\circ\) to the horizontal.
33. A ball is projected with a velocity of \(40 \text{ m/s}\) at an angle of \(60^\circ\) with the horizontal.
- Find the maximum height reached by the ball.
- Find the horizontal range.
(Take \(g = 10 \text{ m/s}^2\))
Solution:
Given:
- Initial velocity (\(u\)) = \(40 \text{ m/s}\)
- Angle of projection (\(\theta\)) = \(60^\circ\)
- Acceleration due to gravity (\(g\)) = \(10 \text{ m/s}^2\)
1. Maximum Height
Using:
\(h = \frac{u^2 \sin^2 \theta}{2g}\)
\(h = \frac{40^2 \times (\sin 60^\circ)^2}{2 \times 10}\)
\(h = \frac{1600 \times \frac{3}{4}}{20} = 60 \text{ m}\)
2. Horizontal Range
Using:
\(X = \frac{u^2 \sin 2\theta}{g}\)
\(X = \frac{40^2 \sin(120^\circ)}{10}\)
\(X = \frac{1600 \times \frac{\sqrt{3}}{2}}{10} = 80\sqrt{3} \text{ m}\)
Final Answer: Maximum height = \(60 \text{ m}\), Range = \(80\sqrt{3} \text{ m}\).
34. In a soccer practice session the football is kept at the centre of the field 40 yards from the 10 ft high goalposts. A goal is attempted by kicking the football at a speed of \(64 \text{ ft/s}\) at an angle of \(45^\circ\) to the horizontal. Will the ball reach the goal post ?
(Take \(g = 32.2 \text{ ft/s}^2\))
Solution:
1. Given Values
- Horizontal distance (\(x\)) = \(40 \text{ yd} = 120 \text{ ft}\)
- Initial velocity (\(u\)) = \(64 \text{ ft/s}\)
- Angle (\(\theta\)) = \(45^\circ\)
- Acceleration (\(g\)) = \(32.2 \text{ ft/s}^2\)
2. Time to reach the goal post
Using:
\(x = u \cos\theta \cdot t\)
\(t = \frac{120}{64 \cos 45^\circ} \approx 2.65 \text{ s}\)
3. Height of the ball at this time
Using:
\(y = u \sin\theta \cdot t – \frac{1}{2}gt^2\)
\(y = 64 \sin 45^\circ (2.65) – \frac{1}{2}(32.2)(2.65)^2\)
\(y \approx 7.08 \text{ ft}\)
4. Conclusion
Since the height of the ball \(7.08 \text{ ft}\) is less than the goal post height \(10 \text{ ft}\), the ball will pass below the top bar and enter the goal.
Final Answer: Yes, the ball will reach the goal post.
35. A popular game in Indian villages is goli which is played with small glass balls called golis. The goli of one player is situated at a distance of $2.0\text{ m}$ from the goli of the second player. This second player has to project his goli by keeping the thumb of the left hand at the place of his goli, holding the goli between his two middle fingers and making the throw. If the projected goli hits the goli of the first player, the second player wins. If the height from which the goli is projected is $19.6\text{ cm}$ from the ground and the goli is to be projected horizontally, with what speed should it be projected so that it directly hits the stationary goli without falling on the ground earlier?
(Take \(g = 9.8 \text{ m/s}^2\))
Solution:
1. Time to reach the ground
For horizontal projection, time depends only on height:
\(t = \sqrt{\frac{2h}{g}}\)
\(t = \sqrt{\frac{2 \times 0.196}{9.8}} = \sqrt{0.04} = 0.2 \text{ s}\)
2. Horizontal velocity
Using:
\(x = ut\)
\(2 = u \times 0.2\)
\(u = \frac{2}{0.2} = 10 \text{ m/s}\)
Final Answer: The initial horizontal velocity is \(10 \text{ m/s}\).
36. Figure (3-E8) shows a \(11.7\text{ ft}\) wide ditch with the approach roads at an angle of \(15^\circ\) with the horizontal. With what minimum speed should a motorbike be moving on the road so that it safely crosses the ditch? Assume that the length of the bike is \(5\text{ ft}\), and it leaves the road when the front part runs out of the approach road.
Solution:
To safely cross the ditch, the motorbike is treated as a projectile. The front end leaves the edge and the bike must completely clear the ditch.
1. Given values
- Ditch width: \(w = 11.7\text{ ft}\)
- Bike length: \(L = 5\text{ ft}\)
- Total horizontal distance: \(R = 11.7 + 5 = 16.7\text{ ft}\)
- Angle of projection: \(\theta = 15^\circ\)
- Acceleration due to gravity: \(g = 32.2\text{ ft/s}^2\)
2. Range formula
\(R = \frac{u^2 \sin(2\theta)}{g}\)
3. Solving for velocity
\(u^2 = \frac{R g}{\sin(2\theta)}\)
\(u^2 = \frac{16.7 \times 32.2}{\sin 30^\circ}\)
\(u^2 = \frac{537.74}{0.5} = 1075.48\)
\(u = \sqrt{1075.48} \approx 32.79\text{ ft/s}\)
4. Final Answer
The minimum speed required is \(32.8\text{ ft/s}\) (approximately \(22.36\text{ mph}\)).
37. A person standing on the top of a cliff \(171\text{ ft}\) high has to throw a packet to his friend standing on the ground \(228\text{ ft}\) horizontally away. If he throws the packet directly aiming at the friend with a speed of \(15.0\text{ ft/s}\), how short will the packet fall?
Solution:
The packet is thrown with initial speed \(u = 15.0\text{ ft/s}\), aimed along the line joining the two points.
1. Angle of projection
\(\tan \theta = \frac{171}{228} = \frac{3}{4}\)
Hence:
- \(\cos \theta = \frac{4}{5}\)
- \(\sin \theta = \frac{3}{5}\)
Velocity components:
- \(u_x = 15 \times \frac{4}{5} = 12\text{ ft/s}\)
- \(u_y = -15 \times \frac{3}{5} = -9\text{ ft/s}\)
2. Time of flight
Using vertical motion:
\(-171 = -9t – 16.1t^2\)
\(16.1t^2 + 9t – 171 = 0\)
Solving:
\(t \approx 3\text{ s}\)
3. Horizontal distance travelled
\(x’ = u_x t = 12 \times 3 = 36\text{ ft}\)
4. Shortfall
\(228 – 36 = 192\text{ ft}\)
Final Answer: The packet falls short by \(192\text{ ft}\).
38. A ball is projected from a point on the floor with a speed of \(15\text{ m/s}\) at an angle of \(60^\circ\) with the horizontal. Will it hit a vertical wall \(5\text{ m}\) away from the point of projection without hitting the floor? Will the answer differ if the wall is \(22\text{ m}\) away?
Solution
Given:
- Initial speed: \(u = 15\text{ m/s}\)
- Angle: \(\theta = 60^\circ\)
- Acceleration due to gravity: \(g = 9.8\text{ m/s}^2\)
- \(\cos 60^\circ = 0.5,\; \tan 60^\circ = \sqrt{3}\)
Trajectory equation:
\(y = x \tan \theta – \frac{g x^2}{2u^2 \cos^2 \theta}\)
Case A: Wall at \(x = 5\text{ m}\)
\(y = 5(\tan 60^\circ) – \frac{9.8 \times 25}{2 \times 15^2 \times 0.25}\)
\(y = 8.66 – 2.18 = 6.48\text{ m}\)
Since \(y > 0\), the ball hits the wall.
Case A Result: The ball hits the wall at a height of \(6.48\text{ m}\).
Case B: Wall at \(x = 22\text{ m}\)
First, find the range:
\(R = \frac{u^2 \sin 2\theta}{g} = \frac{225 \times \sin 120^\circ}{9.8}\)
\(R \approx 19.88\text{ m}\)
Since \(22\text{ m} > 19.88\text{ m}\), the ball lands before reaching the wall.
Case B Result: The ball does not hit the wall.
Final Answer:
- At 5 m: Yes, it hits the wall.
- At 22 m: No, it falls short.
39. Find the average velocity of a projectile between the instants it crosses half the maximum height. It is projected with a speed \(u\) at an angle \(\theta\) with the horizontal.
Solution:
1. Definition of average velocity
\(\vec{v}_{avg} = \frac{\text{Total Displacement}}{\text{Total Time}}\)
For projectile motion between two points at the same height:
- Net vertical displacement is zero
- Horizontal velocity remains constant
2. Vertical component
Since initial and final points are at the same height:
\(\Delta y = 0 \Rightarrow v_{avg,y} = \frac{0}{\Delta t} = 0\)
3. Horizontal component
The horizontal velocity is constant:
\(v_{avg,x} = u \cos \theta\)
4. Resultant average velocity
\(\vec{v}_{avg} = \sqrt{(u\cos\theta)^2 + 0^2}\)
\(\vec{v}_{avg} = u \cos \theta\)
Final Answer:
The average velocity is \(u \cos \theta\) in the horizontal direction.
40. A bomb is dropped from a plane flying horizontally with uniform speed. Show that the bomb will explode vertically below the plane. Is the statement true if the plane flies with uniform speed but not horizontally?
Solution:
1. Independence of motion
Initial horizontal velocity of bomb = u (same as plane)
Horizontal acceleration = 0
Plane horizontal position:
\[
x_p = ut
\]
Bomb horizontal position:
\[
x_b = ut
\]
⇒ x_p = x_b
2. Result
Bomb remains vertically below the plane at all instants.
3. Non-horizontal uniform motion
Plane velocity at angle \(\theta\):
Horizontal component:
\[
v_x = v \cos\theta
\]
Horizontal displacement of both plane and bomb:
\[
x = (v \cos\theta)t
\]
4. Conclusion
Bomb remains vertically below plane for any uniform motion (horizontal or inclined), neglecting air resistance.
41. A boy standing on a long railroad car throws a ball straight upwards. The car is moving on the horizontal road with an acceleration of 1 m/s 2 and the projection velocity in the vertical direction is 9.8 m/s. How far behind the boy will the ball fall on the car ?
Solution:
Given:
- Horizontal acceleration of truck (\(a\)) = \(1 \text{ m/s}^2\)
- Initial vertical velocity (\(u_y\)) = \(9.8 \text{ m/s}\)
- Acceleration due to gravity (\(g\)) = \(9.8 \text{ m/s}^2\)
1. Time of flight
Since the ball is thrown vertically:
\(t = \frac{2u_y}{g} = \frac{2 \times 9.8}{9.8} = 2 \text{ s}\)
2. Horizontal motion
- Ball: retains initial horizontal velocity of the truck →
\(S_b = ut\) - Truck: continues to accelerate →
\(S_c = ut + \frac{1}{2}at^2\)
3. Separation distance
\(\Delta S = S_c – S_b = \frac{1}{2}at^2\)
\(\Delta S = \frac{1}{2} \times 1 \times (2)^2 = 2 \text{ m}\)
Final Answer: The ball lands \(2 \text{ m}\) behind the person.
42. A ball is projected horizontally from the top of a staircase. Each step is \(20 \text{ cm}\) wide and \(20 \text{ cm}\) high. If the ball is projected from the origin (point \(A\)), find the minimum horizontal velocity required for the ball to just touch the edge of the second step (point \(E\)) and reach the ground.
Solution:
1. Coordinates of point \(E\)
- Horizontal distance: \(x = 20 + 20 = 40 \text{ cm}\)
- Vertical distance: \(y = -20 \text{ cm}\)
2. Given
- Angle of projection (\(\theta\)) = \(0^\circ\)
- Acceleration due to gravity (\(g\)) = \(1000 \text{ cm/s}^2\)
3. Trajectory equation
\(y = x \tan \theta – \frac{g x^2 \sec^2 \theta}{2u^2}\)
For \(\theta = 0^\circ\):
\(y = -\frac{g x^2}{2u^2}\)
4. Substitution
\(-20 = -\frac{1000 \times (40)^2}{2u^2}\)
\(20 = \frac{1000 \times 1600}{2u^2}\)
\(40u^2 = 1{,}600{,}000\)
\(u^2 = 40{,}000\)
\(u = 200 \text{ cm/s} = 2 \text{ m/s}\)
Final Answer: The minimum horizontal velocity is \(2 \text{ m/s}\).
43. A person is standing on a truck moving with a constant velocity of 14.7 m/s on a horizontal road. The man throws a ball so that it returns to the truck after the truck has moved 58.8 m. Find the speed and angle of projection (a) as seen from the truck, (b) as seen from the road.
Solution:
1. Time of flight
Truck speed = 14.7 m/s, distance = 58.8 m
\[
T = \frac{58.8}{14.7} = 4 \text{ s}
\]
Ball time of flight = 4 s
2. (a) From truck frame
Horizontal velocity = 0 (relative to truck)
Vertical motion:
\[
T = \frac{2u_y}{g}
\]
\[
4 = \frac{2u_y}{9.8} \Rightarrow u_y = 19.6 \text{ m/s}
\]
Result (a): Speed = 19.6 m/s, Angle = 90°
3. (b) From ground frame
Horizontal velocity:
\(u_x = 14.7 \, \text{m/s}\)
Vertical velocity:
\(u_y = 19.6 \, \text{m/s}\)
Resultant speed:
\[
u = \sqrt{14.7^2 + 19.6^2} = 24.5 \text{ m/s}
\]
Angle of projection:
\[
\tan\theta = \frac{19.6}{14.7} = \frac{4}{3}
\]
\[
\theta \approx 53^\circ
\]
Result (b): Speed = 24.5 m/s, Angle = 53°
Final Answer:
(a) 19.6 m/s, 90°
(b) 24.5 m/s, 53°
44. The benches of a cricket stadium are 1 m wide and 1 m high. A batsman hits a ball from a height of 1 m with speed 35 m/s at angle 53°. First bench is 110 m away. Find on which bench the ball will land.
Solution:
1. Velocity components
\[
u_x = 35\cos 53^\circ \approx 21 \text{ m/s}
\]
\[
u_y = 35\sin 53^\circ \approx 28 \text{ m/s}
\]
2. Trajectory equation
\[
y = x\tan\theta – \frac{gx^2}{2u^2\cos^2\theta}
\]
Substitute values:
\[
y \approx \frac{4x}{3} – \frac{x^2}{90}
\]
3. Bench line
\[
y = x – 110
\]
4. Intersection
\[
x – 110 = \frac{4x}{3} – \frac{x^2}{90}
\]
\[
x^2 – 30x – 9900 = 0
\]
\[
x \approx 115.6 \text{ m}
\]
5. Bench identification
Bench ranges:
Bench 1: 110–111 m
Bench 2: 111–112 m
Since \(x \approx 115.6\), it lies in:
Bench 6
Final Answer: The ball hits the 6th bench.
45. A man is sitting on the shore of a river. He is in the line of a 1.0 m long boat and is 5.5 m away from the centre of the boat. He throws an apple with speed 10 m/s. Find minimum and maximum angles of projection for a successful shot. (Both points at same level).
Solution:
1. Given data
\[
u = 10 \,\text{m/s}, \quad g = 10 \,\text{m/s}^2
\]
Boat length = 1 m
Distance of centre = 5.5 m
2. Required range
Near edge:
\[
R_1 = 5.5 – 0.5 = 5.0 \,\text{m}
\]
Far edge:
\[
R_2 = 5.5 + 0.5 = 6.0 \,\text{m}
\]
So,
5 ≤ R ≤ 6 m
3. Range equation
\[
R = \frac{u^2 \sin 2\theta}{g}
\]
Substitute values:
\[
R = \frac{100 \sin 2\theta}{10} = 10 \sin 2\theta
\]
4. For R = 5 m
\[
5 = 10 \sin 2\theta \Rightarrow \sin 2\theta = 0.5
\]
\[
2\theta = 30^\circ \text{ or } 150^\circ
\]
\[
\theta = 15^\circ \text{ or } 75^\circ
\]
5. For R = 6 m
\[
6 = 10 \sin 2\theta \Rightarrow \sin 2\theta = 0.6
\]
\[
2\theta \approx 37^\circ \text{ or } 143^\circ
\]
\[
\theta \approx 18.5^\circ \text{ or } 71.5^\circ
\]
6. Final range of angles
To land inside the boat:
Low-angle range: \(\;15^\circ \le \theta \le 18.5^\circ\)
High-angle range: \(\;71.5^\circ \le \theta \le 75^\circ\)
Final Answer:
Minimum angle = 15°
Maximum angle = 75°
46. A river 400 m wide is flowing at a rate of 2.0 m/s. A boat is sailing at a velocity of 10 m/s with respect to the water, in a direction perpendicular to the river. (a) Find the time taken by the boat to reach the opposite bank. (b) How far from the point directly opposite to the starting point does the boat reach the opposite bank?
Solution:
1. Given Data
\[
d = 400\,\text{m}, \quad v_r = 2.0\,\text{m/s}, \quad v_{bw} = 10\,\text{m/s}
\]
(a) Time taken to reach the opposite bank
\[
t = \frac{d}{v_{bw}} = \frac{400}{10} = 40\,\text{s}
\]
Time = 40 s
(b) Distance from the point directly opposite (drift)
\[
x = v_r \cdot t = 2.0 \times 40 = 80\,\text{m}
\]
Drift = 80 m
Final Answer:
(a) 40 s
(b) 80 m downstream
47. A swimmer wishes to cross a 500 m wide river flowing at 5 km/h. His speed with respect to water is 3 km/h. (a) If he heads in a direction making an angle θ with the flow, find the time he takes to cross the river. (b) Find the shortest possible time to cross the river.
Solution:
1. Given Data
\[
d = 500\,\text{m} = 0.5\,\text{km}, \quad v_r = 5\,\text{km/h}, \quad v_s = 3\,\text{km/h}
\]
(a) Time to cross
Perpendicular component of swimmer’s velocity:
\[
v_y = v_s \sin\theta = 3\sin\theta
\]
Time taken:
\[
t = \frac{0.5}{3\sin\theta}\,\text{hours}
\]
Convert to minutes:
\[
t = \frac{0.5}{3\sin\theta} \times 60
\]
\[
t = \frac{30}{3\sin\theta} = \frac{10}{\sin\theta}\,\text{minutes}
\]
Time = 10 / sinθ minutes
(b) Minimum time
Minimum time occurs when:
\[
\sin\theta = 1
\]
(i.e., swimmer goes perpendicular to flow)
\[
t_{\min} = \frac{10}{1} = 10\,\text{minutes}
\]
Final Answer:
(a) \( \frac{10}{\sin\theta} \) minutes
(b) 10 minutes
48. Consider the situation of the previous problem. The man has to reach the other shore at the point directly opposite to his starting point. If he reaches the other shore somewhere else, he has to walk down to this point. Find the minimum distance that he has to walk.
Solution:
1. Given data
\[
d = 500\,\text{m} = 0.5\,\text{km}, \quad v_r = 5\,\text{km/h}, \quad v_s = 3\,\text{km/h}
\]
2. Drift distance
Time to cross:
\[
t = \frac{d}{v_s \sin\theta}
\]
Horizontal velocity:
\[
v_x = v_r + v_s \cos\theta
\]
Drift:
\[
x = (v_r + v_s \cos\theta)\frac{d}{v_s \sin\theta}
\]
\[
x = \frac{d}{v_s}\cdot \frac{v_r + v_s \cos\theta}{\sin\theta}
\]
3. Minimisation condition
\[
\frac{d}{d\theta}\left(\frac{v_r + v_s \cos\theta}{\sin\theta}\right)=0
\]
Result:
\[
\cos\theta = -\frac{v_s}{v_r} = -\frac{3}{5}
\]
\[
\sin\theta = \frac{4}{5}
\]
4. Minimum drift
Substitute values:
\[
x_{\min} = \frac{0.5}{3}\cdot \frac{5 + 3(-3/5)}{4/5}
\]
\[
x_{\min} = \frac{0.5}{3}\cdot \frac{3.2}{0.8}
= \frac{0.5}{3}\cdot 4
\]
\[
x_{\min} = \frac{2}{3}\,\text{km}
\]
Final Answer:
Minimum distance to walk = 2/3 km ≈ 667 m
49. An aeroplane has to go from a point A to another point B, 500 km away due 30° east of north. A wind is blowing due north at a speed of 20 m/s. The air-speed of the plane is 150 m/s. (a) Find the direction in which the pilot should head the plane to reach the point B. (b) Find the time taken by the plane to go from A to B.
Solution:
1. Given data
\[
d = 500\,\text{km} = 5\times 10^5\,\text{m}
\]
\[
v_w = 20\,\text{m/s}, \quad v_{pa} = 150\,\text{m/s}
\]
Direction of AB = 30° east of north
(a) Direction to head the plane
Perpendicular component of wind w.r.t. AB:
\[
v_{w\perp} = 20\sin 30^\circ = 10\,\text{m/s}
\]
Perpendicular component of plane velocity:
\[
v_{pa}\sin\alpha
\]
Condition for no drift:
\[
150\sin\alpha = 10
\]
\[
\sin\alpha = \frac{1}{15}
\]
\[
\alpha = \sin^{-1}\left(\frac{1}{15}\right) \approx 3.82^\circ
\]
Direction: 3.82° east of line AB
or
26.18° east of north
(b) Time taken
Component along AB:
\[
v_{pa\parallel} = 150\cos\alpha \approx 150\sqrt{1-\frac{1}{225}} \approx 149.66\,\text{m/s}
\]
Wind component along AB:
\[
v_{w\parallel} = 20\cos 30^\circ \approx 17.32\,\text{m/s}
\]
Total speed:
\[
v = 149.66 + 17.32 = 166.98\,\text{m/s}
\]
Time:
\[
t = \frac{5\times10^5}{166.98} \approx 2994\,\text{s}
\]
\[
t \approx 50\,\text{minutes}
\]
Final Answer:
(a) 3.82° east of AB (≈ 26.2° east of north)
(b) ≈ 50 minutes
50. Two friends A and B are standing a distance x apart in an open field and wind is blowing from A to B. A beats a drum and B hears the sound t1 time after he sees the event. A and B interchange their positions and the experiment is repeated. This time B hears the drum t2 time after he sees the event. Calculate the velocity of sound in still air v and the velocity of wind u. Neglect the time light takes in travelling between the friends.
Solution:
1. Given
\[
\text{Distance} = x,\quad \text{Speed of sound} = v,\quad \text{Wind speed} = u
\]
2. Case 1 (A → B, wind assists)
\[
t_1 = \frac{x}{v+u}
\]
\[
v + u = \frac{x}{t_1}
\]
3. Case 2 (B → A, wind opposes)
\[
t_2 = \frac{x}{v-u}
\]
\[
v – u = \frac{x}{t_2}
\]
4. Solve for v
\[
(v+u)+(v-u)=\frac{x}{t_1}+\frac{x}{t_2}
\]
\[
2v = x\left(\frac{1}{t_1}+\frac{1}{t_2}\right)
\]
\[
v = \frac{x(t_1+t_2)}{2t_1t_2}
\]
5. Solve for u
\[
(v+u)-(v-u)=\frac{x}{t_1}-\frac{x}{t_2}
\]
\[
2u = x\left(\frac{1}{t_1}-\frac{1}{t_2}\right)
\]
\[
u = \frac{x(t_2-t_1)}{2t_1t_2}
\]
Final Answer:
Velocity of sound: \( \frac{x(t_1+t_2)}{2t_1t_2} \)
Velocity of wind: \( \frac{x(t_2-t_1)}{2t_1t_2} \)
51. Suppose A and B in the previous problem change their positions in such a way that the line joining them becomes perpendicular to the direction of wind while maintaining the separation x. What will be the time lag B finds between seeing and hearing the drum beating by A ?
Solution:
1. Given
\[
\text{Distance} = x,\quad \text{Speed of sound} = v,\quad \text{Wind speed} = u
\]
Wind is perpendicular to AB.
2. Effective velocity of sound
To cancel wind component:
\[
v \sin\phi = u
\]
Along AB:
\[
V_{\text{eff}} = v \cos\phi
\]
\[
V_{\text{eff}} = v\sqrt{1-\sin^2\phi}
\]
\[
V_{\text{eff}} = \sqrt{v^2-u^2}
\]
3. Time lag
\[
t = \frac{x}{\sqrt{v^2-u^2}}
\]
4. Using results from Question 50
\[
(v+u)=\frac{x}{t_1}, \quad (v-u)=\frac{x}{t_2}
\]
\[
v^2-u^2 = (v+u)(v-u)
\]
\[
v^2-u^2 = \frac{x^2}{t_1 t_2}
\]
5. Substitute
\[
t = \frac{x}{\sqrt{\frac{x^2}{t_1 t_2}}}
\]
\[
t = \frac{x}{\frac{x}{\sqrt{t_1 t_2}}}
\]
\[
t = \sqrt{t_1 t_2}
\]
Final Answer:
Time lag = √(t₁ t₂)
52. Six particles situated at the corners of a regular hexagon of side a move at a constant speed v. Each particle maintains a direction towards the particle at the next corner. Calculate the time the particles will take to meet each other.
Solution:
1. Nature of motion
Due to symmetry, all particles meet at the centre of the hexagon.
Initial separation between adjacent particles = a
2. Relative motion of adjacent particles
Speed of each particle = v
Angle between velocity of one particle and the line joining adjacent particle = 60°
Approach component of second particle along the line:
\[
v \cos 60^\circ = \frac{v}{2}
\]
3. Velocity of approach
\[
v_{app} = v – \frac{v}{2} = \frac{v}{2}
\]
4. Time to meet
\[
t = \frac{\text{initial separation}}{\text{relative speed}}
\]
\[
t = \frac{a}{v/2}
\]
\[
t = \frac{2a}{v}
\]
Final Answer:
\[
t = \frac{2a}{v}
\]