Question

A very broad elevator is going up vertically with a constant acceleration of 2$m{{s}^{-1}}$. At the instant when its velocity is 4 m/s, a ball is projected from the floor of the lift with a speed of 16 m/s relative to the ground at an elevator of ${{30}^{\circ }}$. The time taken by the ball to return the floor is
(g=10$m{{s}^{-2}}$)
$\text{A}\text{. }\dfrac{3}{2}s$
$\text{B}\text{. }\dfrac{2}{3}s$
$\text{C}\text{. }\dfrac{5}{4}s$
$\text{D}\text{. }\dfrac{4}{5}s$

Answer 1

Calculate the relative acceleration of the ball with respect to the elevator. Then calculate the initial velocity of the ball in the vertical direction with respect to the elevator. Using the formula for the time of flight, calculate the time taken for the ball to return to the floor.
Formula used:
${{a}_{B,E}}={{a}_{B}}-{{a}_{E}}$
${{v}_{B,E}}={{v}_{B}}-{{v}_{E}}$
$T=\dfrac{2u\sin \theta }{g}$

Complete answer:
It is given that a ball is projected inside a moving elevator. Since the floor of the elevator is moving upwards, the time taken for the ball to hit the floor will be less than that of when the ball is projected on the ground.
Let us analyse the case with respect to the elevator. We know that when the ball is projected, due to the gravitational force acting downwards the ball will accelerate down. The acceleration of the ball will be –g = -10$m{{s}^{-2}}$ (i.e. acceleration due to gravity).
But the elevator is accelerating in the upward direction with an acceleration of 2$m{{s}^{-2}}$.
Relative acceleration of the ball with respect to the elevator will be ${{a}_{B,E}}={{a}_{B}}-{{a}_{E}}$.
Here, ${{a}_{B}}=-10m{{s}^{-2}}$ and ${{a}_{E}}=2m{{s}^{-2}}$
$\Rightarrow {{a}_{B,E}}=-10-2=-12m{{s}^{-2}}$.
This means that with respect to the elevator, the ball is accelerating downwards with an acceleration of $12m{{s}^{-2}}$.
When the ball is projected, it has some horizontal and some vertical velocity. AT the time of projection, the horizontal velocity of the ball is $16\cos {{30}^{\circ }}=16\times \dfrac{\sqrt{3}}{2}=8\sqrt{3}m{{s}^{-1}}$.
And its vertical velocity is $16\sin {{30}^{\circ }}=16\times \dfrac{1}{2}=8m{{s}^{-1}}$.
It is said that at this time, the velocity of the elevator is $4m{{s}^{-1}}$ upwards.
Since the elevator is stationary in the horizontal direction, the horizontal velocity of the ball with respect to the elevator is the same as that of with respect to the ground.
However, the vertical relative velocity will be different.
The initially relative velocity of the ball with respect to the elevator will be ${{v}_{B,E}}={{v}_{B}}-{{v}_{E}}$.
Here, ${{v}_{B}}=8m{{s}^{-1}}$ and ${{v}_{B}}=4m{{s}^{-1}}$.
${{v}_{B,E}}=8-4=4m{{s}^{-1}}$.
This means that with respect to the elevator, the ball projected with a velocity such that its velocity in the horizontal direction is $8\sqrt{3}m{{s}^{-1}}$ and its velocity in its vertical direction is $4m{{s}^{-1}}$.
We know that the time of flight of a projectile projected on the ground is given as $T=\dfrac{2u\sin \theta }{g}$….(i), where, $u\sin \theta$ is the initial velocity of the projectile in the vertical direction and g is the acceleration due to gravity.
With rest to the elevator, the formula will remain the same but the values will be different. In this case, $u\sin \theta =4m{{s}^{-1}}$ and $g=12m{{s}^{-2}}$.
Substitute The values in (i).
$T=\dfrac{2(4)}{(12)}=\dfrac{2}{3}s$.
This means that the ball will return to the floor after $\dfrac{2}{3}s$.

Hence, the correct option is B.

Note:
Remember that when we use relative motions of other objects with respect to an object, the velocities, displacements and accelerations of the object may be different from the actual quantities but the time intervals are the same irrespective of the relative motion.