Question

A cricket ball thrown across a field is at heights ${{h}_{1}}$ and ${{h}_{2}}$from the point of projection at times ${{t}_{1}}$ and ${{t}_{2}}$ respectively after the throw. The ball is caught by a fielder at the same height as that of projection. The time of flight of the ball in this journey is:
$\text{A}\text{. }\dfrac{{{h}_{1}}{{t}_{2}}^{2}-{{h}_{2}}{{t}_{1}}^{2}}{{{h}_{1}}{{t}_{2}}-{{h}_{2}}{{t}_{1}}}$
$\text{B}\text{. }\dfrac{{{h}_{1}}{{t}_{1}}^{2}+{{h}_{2}}{{t}_{2}}^{2}}{{{h}_{1}}{{t}_{2}}+{{h}_{2}}{{t}_{1}}}$
$\text{C}\text{. }\dfrac{{{h}_{1}}{{t}_{2}}^{2}-{{h}_{2}}{{t}_{1}}^{2}}{{{h}_{1}}{{t}_{2}}+{{h}_{2}}{{t}_{1}}}$
$\text{D}\text{. }\dfrac{{{h}_{1}}{{t}_{1}}^{2}-{{h}_{2}}{{t}_{2}}^{2}}{{{h}_{1}}{{t}_{1}}-{{h}_{2}}{{t}_{2}}}$

Answer 1

Hint: We will apply the equation of motion to find the relation between height ${{h}_{1}},{{h}_{2}}$ and time ${{t}_{1}},{{t}_{2}}$. Time of flight depends on the initial velocity of the object and the angle of projection. The value of time of flight can be obtained by using the direct formula and putting the required values.

Formula used:
\[T=\dfrac{2u\sin \theta }{g}\]

Complete step by step answer:
When an object or particle is projected near the Earth’s surface and it moves alongside a curved path under the action of gravitational force only, then the object is said to be in projectile motion. A projectile is a body upon which the only force acting is gravity. Gravity acts to affect the vertical motion of the projectile, hence causing a vertical acceleration. While the horizontal motion of the projectile is the result of the tendency of an object in motion to remain in motion at a constant velocity.
The time of flight of the projectile motion is the time for which object is in motion, that is, from when the object is projected to the time when it reaches the surface. The time of flight of projectile motion depends on the initial velocity of the object and the angle of projection.
For vertical movement of the ball,
\[{{h}_{1}}=u\sin \theta {{t}_{2}}-\dfrac{1}{2}g{{t}_{2}}^{2}\]
Or, ${{t}_{1}}=\dfrac{{{h}_{2}}+\dfrac{1}{2}g{{t}_{1}}^{2}}{u\sin \theta }$
Similarly,
${{h}_{2}}=u\sin \theta {{t}_{2}}-\dfrac{1}{2}g{{t}_{2}}^{2}$
Or, ${{t}_{2}}=\dfrac{{{h}_{2}}+\dfrac{1}{2}g{{t}_{2}}^{2}}{u\sin \theta }$
From above equations,
$\dfrac{{{t}_{1}}}{{{t}_{2}}}=\dfrac{\dfrac{{{h}_{1}}+\dfrac{1}{2}g{{t}_{1}}^{2}}{u\sin \theta }}{\dfrac{{{h}_{2}}+\dfrac{1}{2}g{{t}_{2}}^{2}}{u\sin \theta }}$
${{h}_{1}}{{t}_{2}}-{{h}_{2}}{{t}_{1}}=\dfrac{g}{2}\left( {{t}_{1}}{{t}_{2}}^{2}-{{t}_{1}}^{2}{{t}_{2}} \right)$
The time of flight of the cricket ball,
\[T=\dfrac{2u\sin \theta }{g}=\dfrac{2}{g}\left( u\sin \theta \right)\]
Where,
$T$ is the time of flight
$u$ is the initial velocity of projectile
$\theta $ is the angle of the projection
$g$ is acceleration due to gravity
$\begin{align}
  & T=\dfrac{2}{g}\left[ \dfrac{{{h}_{1}}+\dfrac{1}{2}g{{t}_{1}}^{2}}{{{t}_{1}}} \right]=\dfrac{2}{g}\left[ \dfrac{{{h}_{1}}}{g}+\dfrac{{{t}_{1}}^{2}}{2} \right] \\
 & \dfrac{{{h}_{1}}}{{{t}_{1}}}\times \dfrac{2}{g}+{{t}_{1}} \\
 & =\dfrac{{{h}_{1}}}{{{t}_{1}}}\times \left( \dfrac{{{t}_{1}}{{t}_{2}}^{2}-{{t}_{1}}^{2}{{t}_{2}}}{{{h}_{1}}{{t}_{2}}-{{h}_{2}}{{t}_{1}}} \right)+{{t}_{1}} \\
 & =\dfrac{{{h}_{1}}{{t}_{1}}{{t}_{2}}^{2}-{{h}_{1}}{{t}_{1}}^{2}{{t}_{2}}+{{h}_{1}}{{t}_{1}}^{2}{{t}_{2}}-{{h}_{2}}{{t}_{1}}^{3}}{{{t}_{1}}({{h}_{1}}{{t}_{2}}-{{h}_{2}}{{t}_{1}})} \\
 & =\dfrac{{{h}_{1}}{{t}_{2}}^{2}-{{h}_{2}}{{t}_{1}}^{2}}{{{h}_{1}}{{t}_{2}}-{{h}_{2}}{{t}_{1}}}
\end{align}$
We get,
$T=\dfrac{{{h}_{1}}{{t}_{2}}^{2}-{{h}_{2}}{{t}_{1}}^{2}}{{{h}_{1}}{{t}_{2}}-{{h}_{2}}{{t}_{1}}}$
Time of flight of ball in the journey is$\dfrac{{{h}_{1}}{{t}_{2}}^{2}-{{h}_{2}}{{t}_{1}}^{2}}{{{h}_{1}}{{t}_{2}}-{{h}_{2}}{{t}_{1}}}$
Hence, the correct option is A.

Note: Students should note that the horizontal component of the projectile's velocity does not change during the motion since no force is acting along the horizontal direction. Vertical component of velocity changes due to the force of gravity.