An attacker at the base of a castle wall 3.50 m high throws a rock straight up w

Question

An attacker at the base of a castle wall 3.50 m high throws a rock straight up with speed 9.00 m/s from a height of 1.45 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 9.00 m/s and moving between the same two 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 Yes No e) Explain physically why it does or does not agree.

Explanation / Answer

(A) Applying vf^2 - vi^2 = 2 a d in vertical direction.

v0 = 9 m/s

a = - 9.8 m/s^2

for maximum height, vf = 0

0^2 - 9^2 = 2 (-9.8)(h - 1.45)

h - 1.45 = 4.13 m

h = 5.58 m

so yes, it will reach the top.

(B) v^2 - 9^2 = 2 (-9.8)(3.60 - 1.45)

v = 6.23 m/s

(C) v^2 - 9^2 = 2( 1.45 - 3.60)(-9.8)

v = 11.1 m/s

change in speed = 11.1 - 9 = 2.1 m/s

(D)
for initial case( going upward), change = 9 - 6.23 = 2.77 m/s

so No.

(E) change in speed = time x acceleration

acceleration is same in both case.

but while going upward, it will take longer time so greater change.

while going down,it will take less time hence smaaller change.

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