# Question #2f8c3

Jan 18, 2017

A

#### Explanation:

It is a case of Horizontally Launched Projectile

Let us calculate Time of flight $T$
It is the total time taken by the steel ball to fall through height $h$ and cover distance $r$.

$y = {u}_{y} t + \frac{1}{2} g {t}^{2}$
For $y = h , t = T$, we get
$h = 0 \times T + \frac{1}{2} g {T}^{2}$
$\implies T = \sqrt{\frac{2 h}{g}}$

Range $r$ is the horizontal distance covered during the time of flight $T$.

Using kinematic equation
$x = {u}_{x} t$
When $t = T , x = r$. Therefore we get

$r = {u}_{x} T$
$r = u \times \sqrt{\frac{2 h}{g}}$

$\implies {r}^{2} = k h$ .....(1)
where $k = 2 {u}^{2} / g$ and is a constant

Equation (1) is a quadratic passing through origin. As it is of the form ${y}^{2} = x$, hence option A.