A rocket is fired at a speed of 75.0 m/s from ground level, at an angle of 60.0º above the horizontal. The rocket is fired toward an 15.0 m high wall, which is located 27.0 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 by how much the rocket clears the top of the wall, we can break down the motion into vertical and horizontal components.
First, let's calculate the time it takes for the rocket to reach the wall. We can use the horizontal distance and the horizontal component of the initial velocity.
The horizontal component of the initial velocity can be found using trigonometry:
\(v_x = v \cdot \cos(\theta)\)
where \(v\) is the initial velocity of the rocket (75.0 m/s) and \(\theta\) is the launch angle (60.0º).
\(v_x = 75.0 \cdot \cos(60.0º) = 75.0 \cdot 0.5 = 37.5 \, \text{m/s}\)
Next, we can calculate the time taken to reach the wall by dividing the horizontal distance by the horizontal velocity:
\(t = \frac{d_x}{v_x}\)
where \(d_x\) is the horizontal distance to the wall (27.0 m).
\(t = \frac{27.0}{37.5} \, \text{s}\) (approx. 0.72 s)
Now, let's find out the maximum height reached by the rocket. For this, we need to calculate the vertical component of the initial velocity.
The vertical component of the initial velocity can be found using trigonometry:
\(v_y = v \cdot \sin(\theta)\)
\(v_y = 75.0 \cdot \sin(60.0º) = 75.0 \cdot 0.866 \approx 64.95 \, \text{m/s}\)
Now let's calculate the time taken for the rocket to reach the maximum height. We can use the equation of motion in the vertical direction:
\(v_y = u_y + a_y \cdot t\), where gravity is acting downward (\(a_y = -9.8 \, \text{m/s}^2\)).
At the highest point, the final vertical velocity (\(v_y\)) is 0. So we have:
\(0 = 64.95 - 9.8 \cdot t\)
Solving for \(t\):
\(64.95 = 9.8 \cdot t\)
\(t = \frac{64.95}{9.8} \approx 6.63 \, \text{s}\)
Now, let's find the maximum height (\(y\)) reached by the rocket using the equation of motion in the vertical direction:
\(y = u_y \cdot t + \frac{1}{2} \cdot a_y \cdot t^2\)
\(y = 64.95 \cdot 6.63 + 0.5 \cdot (-9.8) \cdot (6.63)^2 \approx 215.78 \, \text{m}\)
Finally, to find by how much the rocket clears the top of the wall, we subtract the height of the wall (15.0 m) from the maximum height reached by the rocket:
\(clearance = y - \text{height of the wall}\)
\(clearance = 215.78 - 15.0 = 200.78 \, \text{m}\)
Therefore, the rocket clears the top of the wall by approximately 200.78 meters.