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 = 2gs 

Where, 

u = Initial velocity of the stone = 0 m/s 

v = Final velocity of the stone 

s = Height of the stone = 19.6 m 

g = Acceleration due to gravity = 9.8 ms⁻² 

∴ v ² − 0² = 2 × 9.8 × 19.6 

⇒ v ² = 2 × 9.8 × 19.6 = (19.6)² 

⇒ v = 19.6 ms⁻¹ 

Hence, the velocity of the stone just before touching the ground is 19.6 ms⁻¹ .