An attacker at the base of a castle wall 3.70 m high throws a rock straight up w

Question

An attacker at the base of a castle wall 3.70 m high throws a rock straight up with speed 7.50 m/s from a height of 1.60 m above the ground. Will the rock reach the top of the wall? If so, what is its speed at the top? If not, what initial speed must it have to reach the top? Find the change in speed of a rock thrown straight down from the top of the wall at an initial speed of 7.50 m/s and moving between the same two points. 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

by the energy conservrtion

height it travell is 3.7-1.6=2.1m

K.E initial + P.Einitial = K.Efinal + P.E.Final

0.5mvi2 +mgh1 = 0.5mvf2 + mgh2

m will be cancell from all over the eqaution

0.5*7.52 + 9.8*1.6 = 0.5vf2 + 9.8*3.7

Vf = 3.88458 m/s

yes it reach the top of casstle

part b

speed at top

vf =3.8845 m/s

change in speed=7.5-3.8845=3.6155 m/s

part c

now energy equation

by cancelling m

0.5vtop2 +ghtop = 0.5vground2 + ghlow

0.5*7.52 + 9.8*3.7 = 0.5vground2 + 9.8*1.6

vground =9.8695 m/s

change in speed =9.8695-7.5=2.369 m/s

part d

so it is not same

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