An attacker at the base of a castle wall 3.66 m high throws a rock straight up w

Question

An attacker at the base of a castle wall 3.66 m high throws a rock straight up with speed 8.50 m/s at a height of 1.53 m above the ground.

A) What initial speed must it have to reach the top?

B) Find the change in the speed of a rock thrown straight down from the top of the wall at an initial speed of 8.50 m/s and moving between the same two points.

C) 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 why or why not.

Explanation / Answer

(A)

for upward thrown rock

vo = ? m/s

displacement = y = 3.66-1.53 = 2.13 m

final velocity V1 = 0

ay = -9.8 m/s^2

from equations of motion

v^2 - vo^2 = 2*ay*y

0 - v^2 = -2*9.8*2.13

v = 6.46 m/s

B)

vo = -8.5 m/s

y = -2.13

ay = -9.8 m/s^2

v^2-8.5^2 = 2*9.8*2.13

v = -10.7 m/s

change = 10.7-8.5 = 2.2 m/s

C)

change in upward = 6.46 m/s

change in down ward = 2.2 m/s

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