An attacker at the base of a castle wall 3.60 m high throws a rock straight up w

Question

An attacker at the base of a castle wall 3.60 m high throws a rock straight up with speed 8.00 m/s from a height of 1.60 m above the ground. (a) Will the rock reach the top of the wall? Yes No (b) If so, what is its speed at the top? If not, what initial speed must it have to reach the top? m/s (c) Find the change in speed of a rock thrown straight down from the top of the wall at an initial speed of 8.00 m/s and moving between the same two points. m/s (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? Explain physically why it does or does not agree.

Explanation / Answer

(b) Since gravity works at 9.81m/s^2 you would need to calculate the deceleration of the upward movement. For the SUVAT equation: S = 3.54m (Presuming the person is flat, considering his height isn't given) U = -8.30m/s (Taking downwards to be positive) V = --- (We need this) A = 9.81m/s^2 (Due to gravity) T = --- (Unimportant) The equation I picked is: V^2 = U^2 + 2as So V = /-8.30^2 + (2*9.81*3.54) V = 0.75m/s That's what I got. Not sure if it's completely right. Been almost a year since I've done vectors. (c) S = 3.54m U = 8.30m/s V = --- (Again, need this) A = 9.81m/s^2 T = --- (Unimportant) Same equation /8.30^2 + (2*9.81*3.54) = 11.76m/s (d) It does not agree as the rock moving upwards is dragged down by gravity so it's slowed down. While gravity pulls the rock falling downwards down, and thus making it accelerate to a faster speed. The rock moving upwards will take longer to reach the top of the wall than the one from the top to reach the fall.

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