Question

A boy on a cliff 49 m high drops a stone.One second later he drops the second stone after the first both hit the ground at the same time what speed did he throws the second stone

Answer 1

for first stone 

h = 49m

g = 9.8m/s²

t = ?

u = 0

apply 

h = ut + gt²/2

49 = 0 + 9.8t²/2

t² = 10

t = √10 = 3.16s

for second stone

t = 3.16 - 1 = 2.16s

h = 49m

g = 9.8m/s²

u = ?

again apply

h = ut + gt²/2

49  = u*2.16 + 9.8*2.16²/2

49 = 2.16*u + 22.86

26.14/2.16 = u

u = 12.10m/s