A ball is gently dropped from a height of 20 m.

Solution:
Given, initial velocity of ball, $\mathrm{u}=0$

Final velocity of ball, $\mathrm{v}=?$

Distance through which the ball falls, $\mathrm{s}=20 \mathrm{~m}$

Acceleration $\mathrm{a}=10 \mathrm{~ms}^{-2}$

Time of fall, $\mathrm{t}=$ ?

We know

$v^{2}-u^{2}=2 a s$

or $\mathrm{v}^{2}-0=2 \times 10 \times 20=400$ or $\mathrm{v}=20 \mathrm{~ms}^{-1}$

Now using $\mathrm{v}=\mathrm{u}+$ at we have

$20=0+10 \times t$ or $t=2 s$