A man drops a 10 kg rock from the top of a 20 m ladder. what will be its kinetic energy when it reaches the ground ? what will be its speed just before it hits the ground ?
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Hey!
given m=10kg
d=20m
u=0
so,
=v^2=u^2+2gs
=v^2=0+2*10*20
=v^2=400
=v=20m/s
so, K.E=1/2mv^2
=1/2*10*20^2
=1/2*10*400
=2000j
Hope it helps you
given m=10kg
d=20m
u=0
so,
=v^2=u^2+2gs
=v^2=0+2*10*20
=v^2=400
=v=20m/s
so, K.E=1/2mv^2
=1/2*10*20^2
=1/2*10*400
=2000j
Hope it helps you
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