Question

A body projected with velocity $u$ at projection angle $\theta$ has horizontal range. For the same velocity and projection angle, its range on the moon surface will be ${g_{moon}} = {g_{earth}}/6$
(A) $36R$
(B) $\dfrac{R}{{36}}$
(C) $\dfrac{R}{{16}}$
(D) $6R$

Answer 1

The range of an object is inversely proportional to the acceleration due to gravity of the location of projection. We need to compare the range of earth to that of the range on the moon using substitution.

Formula used: In this solution we will be using the following formulae;
$R = \dfrac{{{u^2}\sin 2\theta }}{g}$ where $R$ is the range of a projectile, $u$ is the initial velocity of projection, $\theta$ is the angle of projection (with respect to the horizontal axis) and $g$ is the acceleration due to gravity.

Complete Step-by-Step Solution:
A body is said to be projected at a certain initial velocity $u$ with an angle of projection $\theta$ from the horizontal. The range of this certain projectile is said to be $R$. This is on earth. We are to find the range of the same projectile, projected with the same velocity and angle of projection, on the moon.
First we shall recall the range of a body is given as
$R = \dfrac{{{u^2}\sin 2\theta }}{g}$ where $R$ is the range of a projectile, $u$ is the initial velocity of projection, $\theta$ is the angle of projection (with respect to the horizontal axis) and $g$ is the acceleration due to gravity.
Hence, on the moon, the range would be
${R_m} = \dfrac{{{u^2}\sin 2\theta }}{{{g_m}}}$
But ${g_m} = \dfrac{{{g_{earth}}}}{6}$
Then,
${R_m} = \dfrac{{{u^2}\sin 2\theta }}{{\dfrac{{{g_{earth}}}}{6}}} = \dfrac{{6{u^2}\sin 2\theta }}{{{g_{earth}}}}$
Hence,
${R_m} = 6\left( {\dfrac{{{u^2}\sin 2\theta }}{{{g_{earth}}}}} \right)$
$\Rightarrow {R_m} = 6R$

Hence, the correct option is D.

Note: In actuality, the range would be greater than as estimated. This is because air resistance on earth is high due to the atmosphere of the earth, and this reduces the range from as predicted by the range formula. However on the moon, an atmosphere is almost non-existent, hence, the range is actually very close to as predicted by the range formula. Hence, we can say it is six times the ideal range of that of earth.