Question

A projectile has same range $R$ for two angles of projection. If ${T_1}$ and ${T_2}$ be the same time of flight for the two cases then:
A. ${T_1}{T_2} = R$
B. ${T_1}{T_2} \propto {R^2}$
C. $\dfrac{{{T_1}}}{{{T_2}}} = \tan \theta $
D. $\dfrac{{{T_1}}}{{{T_2}}} = {\tan ^2}\theta $

Answer 1

Hint: We need to use the formulas for the range of a body in projectile motion and the time of flight of a body in projectile motion. Then we need to find the values of the time of flights and then find the product as well as the division result.

Formula Used: The formulae used in the solution are given here.
The range of a projectile is given by,
$R = \dfrac{{{u^2}\sin \left( {2\theta } \right)}}{g}$ where $u$ is the initial velocity of the body in projectile motion, $\theta $ is the angle of projection and $g$ is the acceleration due to gravity.
The time of flight of a body in projectile motion $T = \dfrac{{2u\sin \theta }}{g}$.

Complete step by step answer:
Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory.
Given are two cases where for same speed $u$, the two ranges are equal for two different angles, which happens only when the two throwing angles are $\theta $ and $\left( {{{90}^ \circ } - \theta } \right)$. What does vary in two cases is the time of flight, which has a relation with the angle of projection. So for each angle, we have a time of flight.
The range of a projectile is given by,
$R = \dfrac{{{u^2}\sin \left( {2\theta } \right)}}{g}$
$ \Rightarrow R = \dfrac{{2{u^2}\sin \theta \cos \theta }}{g}$
where $u$ is the initial velocity of the body in projectile motion and $g$ is the acceleration due to gravity.
Here, for angle of projection $\theta $, the time of flight is given by,
${T_1} = \dfrac{{2u\sin \theta }}{g}$
and for angle of projection $\left( {{{90}^ \circ } - \theta } \right)$, time of flight is given by,
${T_2} = \dfrac{{2u\cos \theta }}{g}$
So the product of the two times of flights ${T_1}$ and ${T_2}$, is given by, ${T_1}{T_2} = \dfrac{{4{u^2}\sin \theta \cos \theta }}{g}$.
Thus, ${T_1}{T_2} = 2R$. Hence the product is directly proportional to the range $R$.
Mathematically, ${T_1}{T_2} \propto R$.
Now we divide ${T_1}$ ​by ${T_2}$. Computing the values, we get,
$\dfrac{{{T_1}}}{{{T_2}}} = \dfrac{{\dfrac{{2u\sin \theta }}{g}}}{{\dfrac{{2u\cos \theta }}{g}}} = \tan \theta $.
Hence, Option C is the correct answer.

Note: Air resistance to the motion of the body is to be assumed absent in projectile motion.
In a Projectile Motion, there are two simultaneous independent rectilinear motions. Along the x-axis, there is uniform velocity, responsible for the horizontal (forward) motion of the particle and along the y-axis, there is uniform acceleration, responsible for the vertical (downwards) motion of the particle.
Accelerations in the horizontal projectile motion and vertical projectile motion of a particle: When a particle is projected in the air with some speed, the only force acting on it during its time in the air is the acceleration due to gravity ($g$). This acceleration acts vertically downward. There is no acceleration in the horizontal direction, which means that the velocity of the particle in the horizontal direction remains constant.