Question

Question 14
A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity.

Answer 1

According to the equation of motion under gravity:
$v^2-u^2=2gs$
Where,
u = Initial velocity of the stone
v = Final velocity of the stone
s = Height of the stone
g = Acceleration due togravity
Here u = 0, s = 19.6 m and g = $9.8m{s}^{-2}$
$\therefore v^2-0^2=2\times 9.8\times 19.6$
$v^2=2\times 9.8\times 19.6=384.16$
$v=19.6m{s}^{-1}$
Hence, the velocity of the stone just before touching the ground is $19.6m{s}^{-1}$