Question -
Answer -
Accordingto the equation of motion under gravity:
v2 − u2 = 2 gs
Where,
u = Initial velocity of the stone = 0
v = Final velocity of the stone
s = Height of the stone = 19.6 m
g = Acceleration due to gravity = 9.8 m s−2
∴ v2 − 02 = 2 × 9.8 × 19.6
v2 = 2 × 9.8 × 19.6 =(19.6)2
v = 19.6 m s−1
Hence,the velocity of the stone just before touching the ground is 19.6 m s−1.