Question

A stone is dropped from the top of a 45m high building. How fast will be moving when it reaches the ground?And what will its velocity be?

Answer 1

height , S = 45 m

Initial velocity , U = 0

let final velocity = V

From the laws of kinematics -

$V^2 - U^2 = 2gs$

$\implies V^2 - 0 = 2\times 9.8 \times 45$

$\implies V^2 = 882$

$\implies V = 29.7 m/sec$

It reaches the ground at 29.7 m/sec