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**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.

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