Question

A ball is released from rest at the top of a very tall building neglected air resistance calculate velocity of the ball after 3 seconds

Answer 1

Let the height of the building be H
so by equation of motion

$h = ut + \frac{1}{2} g {t}^{2} \\ t = 3 \: and \: u = 0 \\ so \\ \\ h = \frac{1}{2} 9g \\ \\ now \: we \: have \: to \: find \: velocity \: so \\ {v}^{2} - {u}^{2} = 2gh \\ {v}^{2} = 2g( \frac{1}{2} 9g) \\ {v}^{2} = 9 {g}^{2} \\ v = 3g$