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?​

Answer 1

Answer: 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