Projectile Motion Derivation: The projectile motion refers to the motion of an object that is launched into the air and moves under the influence of gravity. It follows a curved path known as a parabola. To derive the equations of projectile motion, we’ll assume the following:
- The only force acting on the projectile is gravity.
- There is no air resistance.
- The motion occurs in a uniform gravitational field.
Let’s consider a projectile launched at an angle θ with an initial velocity v0. We can break down the initial velocity into its horizontal and vertical components as follows:
Initial horizontal component: v0x = v0 * cos(θ) Initial vertical component: v0y = v0 * sin(θ)
Horizontal Motion: In the horizontal direction, there is no acceleration acting on the projectile, so the velocity remains constant throughout the motion. Therefore, the horizontal displacement (range) can be calculated as:
Horizontal displacement: dx = v0x * t
Vertical Motion: In the vertical direction, the only force acting on the projectile is gravity, which causes acceleration. The acceleration due to gravity is constant and directed downward. Therefore, we can use the equations of motion to analyze the vertical motion.
The vertical displacement (height) can be determined using the equation:
Vertical displacement: dy = v0y * t – (1/2) * g * t^2
where g is the acceleration due to gravity.
The time of flight (the total time the projectile is in the air) can be found by determining the time it takes for the vertical displacement to become zero. Setting dy = 0, we can solve for the time of flight:
0 = v0y * t – (1/2) * g * t^2
This equation is a quadratic equation, and by solving it, we can find the time of flight, denoted as T.
Once we have the time of flight, we can substitute it back into the horizontal displacement equation to find the range:
Range: R = v0x * T
Using these equations, we can calculate various properties of projectile motion, such as the maximum height, time of flight, and range, for a given initial velocity and launch angle.
Note: This derivation assumes a flat surface and negligible effects such as air resistance and the Earth’s rotation. In reality, these factors can affect the actual trajectory of a projectile.
Projectile Motion Calculator
Example: Suppose we have a cannon that launches a projectile at an angle of 45 degrees with an initial velocity of 20 m/s. We want to calculate various properties of the projectile’s motion.
Step 1: Resolve the initial velocity into horizontal and vertical components.
Initial horizontal component: v0x = v0 * cos(θ) = 20 m/s * cos(45°) ≈ 14.14 m/s
Initial vertical component: v0y = v0 * sin(θ) = 20 m/s * sin(45°) ≈ 14.14 m/s
Step 2: Calculate the time of flight.
To find the time of flight, we need to determine when the vertical displacement becomes zero.
0 = v0y * t – (1/2) * g * t^2
Since the initial vertical velocity is 14.14 m/s and the acceleration due to gravity is approximately 9.8 m/s^2, we can solve the quadratic equation:
(1/2) * g * t^2 – v0y * t = 0
By solving this equation, we find that t = 2.87 seconds (approximately). This is the time of flight.
Step 3: Calculate the range.
Now that we have the time of flight, we can substitute it back into the horizontal displacement equation to find the range.
Range: R = v0x * T = 14.14 m/s * 2.87 s ≈ 40.58 m
So, the projectile will travel approximately 40.58 meters horizontally before hitting the ground.
Step 4: Calculate the maximum height.
To find the maximum height reached by the projectile, we can use the vertical displacement equation:
Vertical displacement: dy = v0y * t – (1/2) * g * t^2
Substituting the values, we get:
dy = 14.14 m/s * 2.87 s – (1/2) * 9.8 m/s^2 * (2.87 s)^2 ≈ 20.42 m
Therefore, the maximum height reached by the projectile is approximately 20.42 meters.
In this example, we have determined the time of flight, range, and maximum height of the projectile launched at a 45-degree angle with an initial velocity of 20 m/s. These calculations are based on the assumptions of no air resistance and a uniform gravitational field.
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