A particle is thrown up with velocity $v_{0}$ inside a stationary lift of sufficient height. The time of flight is 'T'. Now it is thrown again with same initial speed $v_{0}$ with respect to lift. If at the time of second thrown, lift is moving up with a velocity $v_{0}$ and uniform acceleration g upward. Which of the following are true?

(a)Displacement-time graph of particle when lift was stationary

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(b)Displacement-time graph of the particle with respect to the moving lift.

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(c)Displacement-time graph of the particle with respect to ground when the lift is moving

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(d) Velocity-time graph of the particle with respect to the moving lift.

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Solution

The correct options are
A
(a)Displacement-time graph of particle when lift was stationary

B
(b)Displacement-time graph of the particle with respect to the moving lift.

C
(c)Displacement-time graph of the particle with respect to ground when the lift is moving

D
(d) Velocity-time graph of the particle with respect to the moving lift.

When the lift was stationary:- velocity of projection = $v_{0}$
Time of flight = T so after T particle comes back so displacement = 0
$S = u * T - \frac{1}{2} g T^{2}$
$t (v_{0} - \frac{1}{2} g T) = 0 \Rightarrow \frac{2 v_{0}}{g} = T$
We know the velocity keeps decreasing, till it becomes 0. So s-t curve will be parabolic.
To find maximum height
$- v_{0}^{2} = - 2 g s s = \frac{v_{o}^{2}}{2 g}$
So graph (a) is correct

When lift is moving:
Lift is moving up with $v_{0}$ , particle is thrown with velocity $v_{0}$ relative to lift i.e. If i am the lift, I will see the particle leaving me with velocity $v_{0} \Rightarrow v_{p / l} = v_{0} . . . . (1)$
Now accerleration of lift, $a_{1} = + g$
Accerleration of particle = $a_{p} = - g$
$a_{p / l} = - g - g = - 2 g . . . . (2)$
Now same laws of motion are applied here:
$s = v_{0} + \frac{1}{2} (- 2 g) t^{2}$ when particle comes back s=0
$0 = v_{0} + - g t^{2}$
$0 = t (v_{0} - g t)$
$t = 0, \frac{v_{0}}{g}$
So particle comes back in $\frac{v_{0}}{g} o r \frac{T}{2}$
From lifts frame, the maximum height will be attained by particle when its velocity w.r.t. lift = 0
$\Rightarrow 0 - (v_{0})^{2} = 2 (- 2 g) s$
$s = \frac{v_{0}^{2}}{4 g} = \frac{S}{2}$

from lifts frame, the maximum height will be attained by particle when its velocity w.r.t. lift =0
$\Rightarrow 0 - (v_{0})^{2} = 2 (- 2 g) s$
$s = \frac{v_{0}^{2}}{4 g} = \frac{S}{2}$

So graph s is correct position time graph of particle w.r.t. the lift.
$\to$ Lift is moving but from grounds from of reference
We know that after being thrown, particle hits lift in $\frac{T}{2} o r \frac{v_{0}}{g}$ seconds
Also, particle was thrown at velocity $v_{0}$ w.r.t. the lift
$i . e . v_{p / l} = v_{0}$
Velocity of lift= $v_{0}$
$\Rightarrow v_{p / l} = v_{p} - v_{l}$
$v_{0} = v_{p} - v_{0}$
$a_{p} = g$ downwards

Particle is caught by lift only after T/2. We need to find the displacement till that time

$s = v_{0} \frac{T}{2} - \frac{1}{2} g (\frac{T}{2})^{2}$
$\frac{v_{0}}{2} \frac{4 v_{0}}{g} - \frac{1}{b} \frac{4 v_{0}}{g}$

$s = \frac{3 v_{0}^{2}}{2 g} = 3 s$
So the graph c is correct

Now,
$v_{p / l} = v_{0}$
$a_{p / l} = - 2 g$
$v_{p / l} = u_{p / l} - 2 g t$
$v_{p / l} = v_{0} - 2 g t$

At, t=0
$v_{p / l} = v_{0}$
Slope is constant i.e. -2g
If $v_{p / l} = v_{0}$
Then, $t^{2} \frac{v_{0}}{2 g} = \frac{T}{4}$

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A particle is thrown up with velocity $v_{0}$ inside a stationary lift of sufficient height. The time of flight is 'T'. Now it is thrown again with same initial speed $v_{0}$ with respect to lift. If at the time of second thrown, lift is moving up with a velocity $v_{0}$ and uniform acceleration g upward. Which of the following are true?