A ball is thrown upwards with an initial velocity of $20m{s}^{-1}$. Find the time of flight and the maximum height reached by the body respectively. ($g=10m{s}^{-2}$)

The correct option is A $4s$
$20m$
Given:
Initial velocity, $u=20m{s}^{-1}$
Acceleration due to gravity, $g=-10m{s}^{-2}$
Upward direction is taken as positive.

At maximum height, velocity will be zero, (v = 0) as the ball will change the direction of motion. Let the maximum height be $H$.
Using third equation of motion,
$v^2=u^2+2as$
$0^2={20}^2+2\times (-10)\times H$
$H=20m$

When the ball returns to the initial point, net displacement is zero. Let the time of flight be $T$. Using second equation of motion, we get:
$s=ut+\frac{1}{2}a{t}^2$
$0=20\times T+\frac{1}{2}\times (-10)\times T^2$
$T=0$ or $T=4s$
Time of flight is more than zero. So, it is 4 s.