Question

A boy on a cliff 49 m high drops a stone. One second later, he throws a stone after a first. They both hit the ground at the same time. with what speed did he throw the second stone?
step by step answer and also show how to find acceleration​

Answer 1

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If time for first stone is T, time for second stone is t−1

s=49m is the same for both stones, u=0 for first stone

v2 =u2+2asv2

=0+2(9.8)(49)

v=31m/s

v=u+at

31=0+9.8t

t=3.16s

t =2.16s

s=ut+0.5at2

49=u ′(2.16)+0.5(9.8)(2.16)²u′

=12.1m/s