A rocket is fired at a speed of 75.0 m/s from ground level, at an angle of 56.3 ° above the horizontal. The rocket is fired toward an 11.0-m high wall, which is located 21.5 m away. The rocket attains its launch speed in a negligibly short period of time, after which its engines shut down and the rocket coasts. 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 down the problem into horizontal and vertical components.
First, let's look at the horizontal component. The rocket is fired at an angle of 56.3° above the horizontal, but after reaching its launch speed, it will only travel horizontally. Therefore, we can analyze the horizontal motion separately.
The time it takes for the rocket to reach the wall can be found using the horizontal distance and the horizontal component of the velocity. The horizontal distance traveled can be calculated using the formula:
distance = velocity * time
In this case, the velocity is the horizontal component of the launch velocity, which can be calculated using trigonometry:
v_horizontal = launch velocity * cos(angle)
Now we can solve for time using the formula:
time = distance / v_horizontal
Substituting the given values:
distance = 21.5 m
launch velocity = 75.0 m/s
angle = 56.3°
v_horizontal = 75.0 m/s * cos(56.3°) ≈ 37.88 m/s
time = 21.5 m / 37.88 m/s ≈ 0.568 s
Next, let's look at the vertical component of the motion. We need to find the height of the rocket when it reaches the wall. We can use the vertical motion equations to solve for this.
The initial vertical velocity is given by:
v_vertical_initial = launch velocity * sin(angle)
Substituting the given values:
v_vertical_initial = 75.0 m/s * sin(56.3°) ≈ 62.79 m/s
The final vertical velocity is zero since the rocket reaches its maximum height and starts falling back down.
Using the equation:
final velocity = initial velocity + (acceleration * time)
where acceleration is -9.8 m/s^2 (due to gravity), we can solve for the time it takes for the rocket to reach its maximum height:
0 = 62.79 m/s - (9.8 m/s^2 * time)
time = 62.79 m/s / 9.8 m/s^2 ≈ 6.41 s
Next, we can calculate the maximum height reached by the rocket using the equation:
max height = initial height + (initial velocity * time) + (0.5 * acceleration * time^2)
Substituting the given values:
initial height = 0 m (ground level)
initial velocity = 62.79 m/s
acceleration = -9.8 m/s^2
time = 6.41 s
max height = 0 m + (62.79 m/s * 6.41 s) + (0.5 * -9.8 m/s^2 * (6.41 s)^2) ≈ 201.3 m
Finally, to find out how much the rocket clears the top of the wall, we subtract the height of the wall from the maximum height reached:
clearance = max height - wall height
Substituting the given values:
wall height = 11.0 m
max height = 201.3 m
clearance = 201.3 m - 11.0 m ≈ 190.3 m
Therefore, the rocket clears the top of the wall by approximately 190.3 meters.