A boy standing on a cliff 480 m high drops a stone. one second later, He throws a second stone after the first. they both hit the ground at the same time. with what speed does he throw the second stone? (take g = 10 metre per second square)
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At first find time taken by rhe first stone to reach the ground by using rhe equation
S=ut+(1/2)at^2
480=0×t+(1/2)9.8×t^2
t=9.89sec
Then use the same equation to find the initial velocity of the second stone.But this time, time should (t-1) since it is thrown 1sec after the first.
S=u(t-1)+(1/2)a(t-1)^2
480=u×8.89+(1/2)9.8×8.89^2
Solving you will get the initial velocity about 10.43 m/sec.
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