Question

A boy on cliff 49m high drops a stone ,one second later he throws another stone vertically downwards.The two stone hit the ground at the same.What was the velocity with which the second stone was thrown?​

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