a stone of mass 3kg is falling from rest from the top ofa hill.what will be its K.E. after 3s.if the hill is 100m high,what will be its K.E.just before hitting the ground.

Asked by ganga kaushik mangena | 27th Feb, 2011, 05:34: AM

Expert Answer:

Dear student
Given mass= 3 kg
u =0
t=3 s
using the equation, v=u+at
v= 10 x 3 = 30m/s
KE after 3 sec = (1/2)x 3 x (30)^2 J
Given h = 100m
Velocity just before striking the ground,
v^2-u^2 = 2ah
v^2 = 2 x 100 x 10=2000
v = m/s
So KE= (1/2)x 3 x (20√5)^2 J

Answered by  | 27th Feb, 2011, 10:16: PM

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