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 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 in 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?

Explanation / Answer

(a) and (b) Let: u be the initial upward speed, v be the speed at height h above the projection level, g be the acceleration due to gravity. v^2 = u^2 - 2gh ...(1) Putting h = 3.65 - 1.55 m, u = 7.40 m/s: v = sqrt[ 7.40^2 - 2 * 9.81(3.65 - 1.55) ] = 3.68 m/s. The rock will reach the top of the wall, and have speed 3.68m/s upwards when it does so. (c) Let: u be the initial projection speed, v be the speed at distance h below. v^2 = u^2 + 2gh ...(2) v = sqrt[ 7.40^2 + 2 * 9.81(3.65 - 1.55) ] = 9.80 m/s. (d) The change in speed for answer (b) is: v - u = 7.40 - 3.68 = 3.72 m/s. The change in speed for answer (c) is: v - u = 9.80 - 7.40 = 2.40 m/s. (d) It is clear from equations (1) and (2) that | v^2 - u^2 | is the same for the upward flight as for the downward. The change in the speed is therefore different. Consider also the equations: v = u - gt (up) v = u + gt (down). | v - u | is the same when the rock is travelling up or down for the same time, not for the same distance. When the rock is travelling down, its speed is always 7.40 m/s or higher. When it is travelling up, its speed is always 7.40 m/s or lower.

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