Question

A stone isreleased from the top of a tower of height 19.6 m. Calculate itsfinal velocity just before touching the ground.

Answer 1

Accordingto the equation of motion under gravity:

v²âˆ’ 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

∴ v²âˆ’ 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 ms^−1_.