A rocket is fired at a speed of 75.0 m/s from ground level, at an angle of 63.0° above the horizontal. The rocket is fired toward an 11.0 m high wall, which is located 22.0 m away. By how much does the rocket clear the top of the wall?
To find out how much the rocket clears the top of the wall, we need to break the rocket's motion into horizontal and vertical components.
Firstly, let's analyze the vertical component of the rocket's motion. We know that the initial vertical velocity is given by the formula:
Viy = V * sin(θ)
where Viy is the initial vertical velocity, V is the initial velocity of the rocket, and θ is the angle of the rocket's motion with respect to the ground.
Substituting the given values, we have:
Viy = 75.0 m/s * sin(63.0°)
Next, we can determine the time it takes for the rocket to reach the wall, using the formula for the time of flight:
t = 2 * Viy / g
where t is the time of flight and g is the acceleration due to gravity (approximately 9.8 m/s²).
Using the calculated Viy value, we can find the time of flight:
t = 2 * (75.0 m/s * sin(63.0°)) / 9.8 m/s²
Now, during this time of flight, the rocket will rise to a maximum height. Using the formula for the maximum height reached by a projectile:
h = (Viy²) / (2 * g)
Substituting the Viy value we calculated earlier, we get:
h = ((75.0 m/s * sin(63.0°))²) / (2 * 9.8 m/s²)
Therefore, the maximum height the rocket reaches is given by this expression.
Now, we know that the wall is 11.0 m high. To find out how much the rocket clears the top of the wall, we need to subtract the height of the wall from the maximum height the rocket reaches.
Let's perform the calculations to find the answer.