from a cliff of 49m high ,a man drops a stone 1s later he throes second stone . they both hit the ground at the same time . find out the speed with which he throws the second stone.

Answer 1

1st Stone:
4.9t^2 = 49
t^2 = 10
Tf = 3.16 s. = Fall time.

2nd stone:
h = Vo*t + 0.5g*t^2 = 49
t = 3.16-1 = 2.16 s. = Fall time.

Vo*2.16 + 4.9*2.16^2 = 49
2.16Vo + 22.86 = 49
2.16Vo = 49-22.86 = 26.14
Vo = 12.1 m/s. = Initial velocity.