An object is projected so that its horizontal range R is maximum. If the maximum height attained by the object is H, then the ratio of Rto H is,
A) 4
_B) $\dfrac {1} {4} $

C) 2
_D) $\dfrac {1} {2} $

Hint: Use the formula of the range $R = \dfrac{{{u^2}\sin 2\theta}} {{2g}}$.

Complete step by step solution:
Let R is the maximum horizontal range of the projectile. The formula of the range is given by:

$R = \dfrac{{{u^2}\sin 2\theta}} {{2g}} $

Where, $R$ is the maximum horizontal range of the projectile, $u$ is the initial velocity, $\theta $is the angle of projection and $g$ is acceleration due to gravity.

For maximum range,

$\sin 2\theta = 1$

Therefore,
$\theta = {45^0} $
Putting the value of $\theta $ in the above equation, we get:
$R = \dfrac{{{u^2}}}{g}$
The formula for the maximum height attained is given by:
$H = \dfrac{{{u^2}{{\sin} ^2}\theta}} {{2g}}$
$H = \dfrac{1}{2} \times \dfrac{{{u^2}}}{g} \times {\sin ^2} (45)$
We know that,
$R = \dfrac{{{u^2}}}{g}$
From above equation,
$H = \dfrac{1}{2} \times R \times {\sin ^2} (45) $
$H = \dfrac{R}{4}$
So, the greatest height attained by the particle is $\dfrac{R} {4} $. Hence, this is the
required solution.
Therefore, ration between $H$ and $R$ is $\dfrac {1} {4} $
Hence, option B is correct.

Note: In a Projectile Motion, there are two simultaneous independent rectilinear motions:
-Along the x-axis: uniform velocity, responsible for the horizontal (forward) motion of the particle.
-Along y-axis: uniform acceleration, responsible for the vertical (downwards) motion of the particle.