For a given velocity, a projectile has the same range $R$ for two angles of projection. If ${t_1}$ and ${t_2}$ are the time of flight in the two cases, then:
A) ${t_1}{t_2}\alpha R$
B) ${t_1}{t_2}\alpha {R^2}$
C) ${t_1}{t_2}\alpha \dfrac{1}{{{R^2}}}$
D) ${t_1}{t_2}\alpha \dfrac{1}{R}$

Hint:The range of the projectile along the horizontal plane is given by $R = \dfrac{{{u^2}\sin 2\theta }}{g}$ . The time of flight for the projectile is given by $t = \dfrac{{2u\sin \theta }}{g}$ . We need to find the angles for which the ranges are equal and then for those ranges we need to find and compare the time period.

Complete step-by-step solution:We need to find two angles for which the range is equal.
The Range $R$ of the projectile motion is given as:
$R = \dfrac{{{u^2}\sin 2\theta }}{g}$
Here, $u$ is the initial velocity at which the projectile is projected.
$\theta $ is the angle of projection
$g$ is the acceleration due to gravity;
We need to find two angles such that their ranges are equal.
Let the two angles be ${\theta _1}$ and ${\theta _2}$ , now the corresponding range will be:
$R = \dfrac{{{u^2}\sin 2{\theta _1}}}{g}$ And $R = \dfrac{{{u^2}\sin 2{\theta _2}}}{g}$
Two ranges are equal when the angles are complementary, that is
${\theta _1} + {\theta _2} = {90^ \circ }$
$ \Rightarrow {\theta _2} = {90^ \circ } - {\theta _1}$ --equation $1$
As time of flight is given as
$t = \dfrac{{2u\sin \theta }}{g}$
The time of flight for angle of projections ${\theta _1}$ and ${\theta _2}$ , will be
${t_1} = \dfrac{{2u\sin {\theta _1}}}{g}$ and ${t_2} = \dfrac{{2u\sin {\theta _2}}}{g}$
From equation $1$ , the time of flight ${t_2}$ , will be
$\;{t_2} = \dfrac{{2u\sin \left( {{{90}^ \circ } - {\theta _1}} \right)}}{g}$
$\; \Rightarrow {t_2} = \dfrac{{2ucos{\theta _1}}}{g}$
We have values of ${t_1},{t_2}$ and we need the relation between ${t_1} \times {t_2}$ and $\;R$ .
$\;{t_1} \times {t_2} = \dfrac{{2u\sin {\theta _1}}}{g} \times \;\dfrac{{2ucos{\theta _1}}}{g}$
Solving further, we get
$\;{t_1} \times {t_2} = 2\dfrac{{{u^2}\left( {2\sin {\theta _1}cos{\theta _1}} \right)}}{{{g^2}}}\;$
$ \Rightarrow \;{t_1} \times {t_2} = 2\dfrac{{{u^2}\sin 2{\theta _1}\;}}{{{g^2}}}$
_{Also, we know that,} $\sin 2{\theta _1} = 2\sin {\theta _1}cos{\theta _1}$
But $R = \dfrac{{{u^2}\sin 2{\theta _1}}}{g}$ , therefore we can have
$\;{t_1} \times {t_2} = 2\dfrac{{{u^2}\sin 2{\theta _1}\;}}{{{g^2}}}$
$\;\theta = {45^ \circ }$
This means that the product of time periods when range is equal is directly proportional to the range.
${t_1}{t_2}\alpha R$

Thus, option A is the correct option.

Note:Remember the formulas for range and time period for a particle in projectile motion. Also, we used some trigonometric identities in this problem, remember those identities too. It is to be noted that the range is maximum when $\;\theta = {45^ \circ }$ and height is maximum when $\;\theta = {0^ \circ }$ that is when a particle is thrown vertically upwards.