Question

a stone is dropped from the top of a 45m high building how fast will it be moving When it reaches the ground and what will its velocity be?

Answer 1

Given

H = 45m

and we take g(acceleration due to gravity )=9.8m/s²

And

V(velocity at ground)

By using energy conservation

or

using Newton's equation of motion

{v²- u² =2as }

here u (initial velocity) = 0

v=v

a = g

s = h

By solving , we get

V = $\sqrt{2gh}$

$\implies$ V= $\sqrt {2\times 9.8\times 45}$

$\implies$ v=$\sqrt{882}$

$\implies$ $\boxed{v=29.698m/s}$ Towards the ground