Question

A projectile can have the same range R for two angles of projection. If ${{t}_{1}}$ and ${{t}_{2}}$ be the times of flight in the two cases, then the product of the two times of flight is directly proportional to
A. $\dfrac{1}{{{R}^{2}}}$
B. $\dfrac{1}{R}$
C. $R$
D. ${{R}^{2}}$

Answer 1

Use the formula for the time of fight and the horizontal range of a projectile for two different angles of projection. Then equate the two ranges and find the relation between the two angles of projection. After this, find the expression for the product of the two times of flight in terms of R.

Formula used:
$T=\dfrac{2u\sin \theta }{g}$
$R=\dfrac{2{{u}^{2}}\sin \theta \cos \theta }{g}$
$\sin {{\theta }_{1}}\cos {{\theta }_{1}}-\sin {{\theta }_{2}}\cos {{\theta }_{2}}=\cos ({{\theta }_{1}}+{{\theta }_{2}})$

Complete step by step answer:
Let us first define some variables for a projectile motion. Suppose a projectile is thrown with an initial velocity ‘u’ in a direction such that the initial velocity of the projectile makes an acute angle $\theta$ with the x axis. Now, the time of flight of the projectile is the time taken by it to hit the ground form the time of launch and it is given as $T=\dfrac{2u\sin \theta }{g}$, where g is the acceleration due to gravity.

The range of the projectile is the horizontal distance covered by it until it hits the ground. The range (R) is given as $R=\dfrac{2{{u}^{2}}\sin \theta \cos \theta }{g}$.
Now, it is given that the projectile has the same range of different angles of projections (say ${{\theta }_{1}}$ and ${{\theta }_{2}}$). The initial speed (u) of the projectile is the same in both cases and ‘g’ is constant near the surface of earth.

Therefore, when the angle of projection is ${{\theta }_{1}}$ the time of flight is ${{t}_{1}}=\dfrac{2u\sin {{\theta }_{1}}}{g}$ …. (i)
and the horizontal range is $R=\dfrac{2{{u}^{2}}\sin {{\theta }_{1}}\cos {{\theta }_{1}}}{g}$ …. (ii)
When the angle of projection is ${{\theta }_{2}}$ the time of flight is ${{t}_{2}}=\dfrac{2u\sin {{\theta }_{2}}}{g}$ ….. (iii)
and the horizontal range is $R=\dfrac{2{{u}^{2}}\sin {{\theta }_{2}}\cos {{\theta }_{2}}}{g}$ ….. (iv)

Now, equate (ii) and (iv).
$\Rightarrow \dfrac{2{{u}^{2}}\sin {{\theta }_{1}}\cos {{\theta }_{1}}}{g}=\dfrac{2{{u}^{2}}\sin {{\theta }_{2}}\cos {{\theta }_{2}}}{g}$
$\Rightarrow \sin {{\theta }_{1}}\cos {{\theta }_{1}}=\sin {{\theta }_{2}}\cos {{\theta }_{2}}$
$\Rightarrow \sin {{\theta }_{1}}\cos {{\theta }_{1}}-\sin {{\theta }_{2}}\cos {{\theta }_{2}}=0$.
But, $\sin {{\theta }_{1}}\cos {{\theta }_{1}}-\sin {{\theta }_{2}}\cos {{\theta }_{2}}=\cos ({{\theta }_{1}}+{{\theta }_{2}})$.
$\Rightarrow \cos ({{\theta }_{1}}+{{\theta }_{2}})=0$
And we know that $\cos {{90}^{\circ }}=0$
$\Rightarrow {{\theta }_{1}}+{{\theta }_{2}}={{90}^{\circ }}$.
$\Rightarrow {{\theta }_{1}}={{90}^{\circ }}-{{\theta }_{2}}$.
Substitute this value in (i).
$\Rightarrow {{t}_{1}}=\dfrac{2u\sin ({{90}^{\circ }}-{{\theta }_{2}})}{g}$
And $\sin ({{90}^{\circ }}-{{\theta }_{2}})=\cos {{\theta }_{2}}$.
Therefore,
$\Rightarrow {{t}_{1}}=\dfrac{2u\cos {{\theta }_{2}}}{g}$ …. (v).
Now, multiply (v) and (iii).
$\Rightarrow {{t}_{1}}{{t}_{2}}=\left( \dfrac{2u\cos {{\theta }_{2}}}{g} \right)\left( \dfrac{2u\sin {{\theta }_{2}}}{g} \right)=\dfrac{4{{u}^{2}}\sin {{\theta }_{2}}\cos {{\theta }_{2}}}{g}$
The above equation can be written as
${{t}_{1}}{{t}_{2}}=\dfrac{2}{g}\left( \dfrac{2{{u}^{2}}\sin {{\theta }_{2}}\cos {{\theta }_{2}}}{g} \right)$.
And $\left( \dfrac{2{{u}^{2}}\sin {{\theta }_{2}}\cos {{\theta }_{2}}}{g} \right)=R$
$\therefore {{t}_{1}}{{t}_{2}}=\dfrac{2}{g}R$.

This means that the products of the two times of flight is directly proportional to R. Hence, the correct option is C.

Note: The term $2\sin \theta \cos \theta$ can be written as $2\sin \theta \cos \theta =\sin 2\theta$.
Therefore, the horizontal range of the projectile can also be written as $R=\dfrac{{{u}^{2}}\sin 2\theta }{g}$.
However, we did not use this formula because if we equate the two ranges that we will get that $\sin 2{{\theta }_{1}}=\sin 2{{\theta }_{2}}$, which then imply that ${{\theta }_{1}}={{\theta }_{2}}$.
But both the angles of projection are different.