Question

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)​

Answer 1

Answer:

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.