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

Hint: 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.