A rocket is fired at a speed of 75.0 m/s from ground level, at an angle of 69.6°

Question

A rocket is fired at a speed of 75.0 m/s from ground level, at an angle of 69.6° above the horizontal. The rocket is fired toward an 11.0 m high wall, which is located 29.0 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?



Airplane flight recorders must be able to survive catastrophic crashes. Therefore, they are typically encased in crash-resistant steel or titanium boxes that are subjected to rigorous testing. One of the tests is an impact shock test, in which the box must survive being thrown at high speeds against a barrier. A 41-kg box is thrown at a speed of 195 m/s and is brought to a halt in a collision that lasts for a time of 5.3 ms. What is the magnitude of the average net force that acts on the box during the collision?

Explanation / Answer

A rocket is fired at a speed of 75.0 m/s from ground level, at an angle of 69.4° above the horizontal. The rocket is fired toward an 11.0 m high wall, which is located 29.0 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 vx = 75.0m/s*cos(69.4) = 26.38 m/s vy = 75.0m/s*sin(69.4) = 70.20m/s distance d = 29m the time to clear the wall is t = d/vx = 29m /26.38 m/s = 1.099 s in the vertical direction, height h change with time as follows: h(t) = vy * t - (1/2) g t2 at t=1.099s, we have: h = 70.20m/s *1.099s -(1/2)*9.8m/s^2(1.099s)^2 = 71.22m Therefore dh = h - 11.0m = 71.22m - 11.0m = 60.22m it clear the top of the wall by 60.22m

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