$A$ stone is dropped from a height of $20\, m$.
$(a)$ How long will it take to reach the ground?
$(b)$ What will be its speed when it hits the ground? $(g=10\, m s^{-2})$

(N/A) Given: height $h = 20\, m$,initial velocity $u = 0$,acceleration due to gravity $g = 10\, m s^{-2}$.
Using the second equation of motion $S = ut + \frac{1}{2}at^2$,where $S = h$ and $a = g$:
$20 = 0 \times t + \frac{1}{2} \times 10 \times t^2$
$20 = 5t^2$
$t^2 = 4$
$t = 2\, s$.
So,it will take $2\, s$ to reach the ground.
$(b)$ Using the first equation of motion $v = u + at$:
$v = 0 + 10 \times 2$
$v = 20\, m s^{-1}$.
Thus,the speed when it hits the ground is $20\, m s^{-1}$.