An attacker at the base of a castle wall 3.65 m high throws a rock straight up w

Question

An attacker at the base of a castle wall 3.65 m high throws a rock straight up with a speed 7.40 m/s from a height of 1.55 m above the ground. (a)Will the rock reach the top of the wall? (b)If so, what is its speed at the top? If not, what initial speed must it have to reach the top? (c)Find the change of speed of a  rock thrown straight down from the top of the wall at an initial speed of 7.40 m/s and moving between the same two points. (d)Does the change in speed of the downward-moving rock agree with the magnitude of the speed change of the rock moving upward between the same elevations? (e)Explain physically why it does or does not agree.

Explanation / Answer

H = 3.65 m, v = 7.40 m/s, h = 1.55 m initial velocity = v (up), displacement = H - h (up), acceleration= g (down) a) find final velocity u u2 - v2 = 2(-g)(H - h) u = [v2 - 2g(H - h)] = 3.69 m/s so the rock will reach the top of the wall. b) u = 3.69 m/s c) initial velocity = v = 7.40 m/s (down), displacement = H - h(down), acceleration = g (down) final velocity u = [v2 + 2g(H - h)] = 9.79m/s change in speed = u - v = 2.09 m/s d) change in speed if moving upward = 3.69 - 7.40 = -3.71 m/s e) no, because acceleration is always down in both cases.

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