Question

A boy on a cliff 49 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 did he throw the second stone ?
(V10 = 3.162)​

Answer 1

Answer:

12.1m/s

Explanation:

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

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

v

2

=u

2

+2as

v

2

=0+2(9.8)(49)

v=31m/s

v=u+at

31=0+9.8t

t=3.16s

t

′

=2.16s

s=ut+0.5at

2

49=u

′

(2.16)+0.5(9.8)(2.16)

2

u

′

=12.1m/s