During a tennis match, a player serves at $23.6{\text{ms}}^{-1}$, the ball leaving the racquet horizontally $2.37\text{m}$ above the court surface. By how much does the ball clear the net which is $12\text{m}$ away and $0.9\text{m}$ high?
(Take $g=10m/s^2$)

The correct option is A $0.17\text{m}$
Let $t$ be the time taken by the ball to clear the net.
Horizontal distance between the racquet and the net $=12\text{m}$
$\Rightarrow 12=u_{x}t=23.6\times t$
Height by which ball will fall in time $t$ is
$y=\frac{1}{2}g{t}^2=\frac{1}{2}\times 10\times {\left(\frac{12}{23.6}\right)}^2=1.3\text{m}$
Let $h$ be the height above the net at which at the ball will pass it.
Then, $2.37=y+h+0.9$
$\therefore h=2.37-0.9-1.3=0.17$ m