Question

A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.

Answer 1

According to 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_.