Question

A shell is fired from a fixed artillery gun, with an initial speed that hits the target on the ground at a distance $R$ from it. ${t_1}$ and ${t_2}$ are the values of the time taken by it to hit the target in two possible ways, the product ${t_1}{t_2}$ is:
(A) $\dfrac{R}{g}$
(B) $\dfrac{R}{{4g}}$
(C) $\dfrac{{2R}}{g}$
(D) $\dfrac{R}{{2g}}$

Answer 1

Since at both times the target is hit, then the range is equal for both times. When one of the angles which can give a particular range is say theta, the other angle which can give the same range would be 90 minus theta.
Formula used: In this solution we will be using the following formulae;
$T = \dfrac{{2u\sin \theta }}{g}$ where $T$ is the time of flight of a projectile, $u$ is the initial velocity of the projectile, $\theta$ is the angle of projection, and $g$ is the acceleration due to gravity.
$R = \dfrac{{{u^2}\sin 2\theta }}{g}$ where $R$ is the range (horizontal displacement from starting point) of the projectile.
$2\sin \theta \cos \theta = \sin 2\theta$ where $\theta$ is an arbitrary angle. $\cos \theta = \sin \left( {90 - \theta } \right)$

Complete Step-by-Step Solution:
Generally, for a particular initial velocity, there are two possible angles of projection which can result in the same range. It is known that if one of the angles is $\theta$, the other would be $90 - \theta$
Now, time of flight of a projectile is given by
$T = \dfrac{{2u\sin \theta }}{g}$ where $T$ is the time of flight of a projectile, $u$ is the initial velocity of the projectile, $\theta$ is the angle of projection, and $g$ is the acceleration due to gravity.
Hence, the time ${t_1}$ and ${t_2}$, will be given as
${t_1} = \dfrac{{2u\sin \theta }}{g}$ and
${t_2} = \dfrac{{2u\sin \left( {90 - \theta } \right)}}{g}$
$\Rightarrow {t_2} = \dfrac{{2u\cos \theta }}{g}$
Hence, the product would be
${t_1}{t_2} = \dfrac{{2u\cos \theta }}{g} \times \dfrac{{2u\sin \theta }}{g}$
$\Rightarrow {t_1}{t_2} = \dfrac{{2{u^2}\left( {2\sin \theta \cos \theta } \right)}}{{{g^2}}}$
From trigonometry, $2\sin \theta \cos \theta = \sin 2\theta$
Thus,
${t_2}{t_2} = \dfrac{{2{u^2}\sin 2\theta }}{{{g^2}}} = \dfrac{2}{g}\left( {\dfrac{{{u^2}\sin 2\theta }}{g}} \right)$
Recall that the range can be given as
$R = \dfrac{{{u^2}\sin 2\theta }}{g}$
Hence,
${t_2}{t_2} = \dfrac{{2R}}{g}$

Hence, the correct option is C

Note: For clarity, the two angles which make up the same range can be gotten from the range equation,
$R = \dfrac{{{u^2}\sin 2\theta }}{g}$
As said above, $\sin 2\theta = 2\cos \theta \sin \theta$
$R = \dfrac{{2{u^2}\cos \theta \sin \theta }}{g}$
But also, from trigonometry,
$\cos \theta = \sin \left( {90 - \theta } \right)$
Hence,
$R = \dfrac{{2{u^2}\sin \left( {90 - \theta } \right)\sin \theta }}{g}$
Hence, the two angles making up the same range are $\sin \left( {90 - \theta } \right){\text{ and }}\sin \theta$
For example when $\theta = 60$ and when $\theta = 90 - 60 = 30$ would give the same range as the products are identical.