#
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)

## Answers

Answered by

0

**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

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