Question

An elevator at rest which is at ${10}^{\text{th}}$ floor of a building is having a plane mirror fixed to its floor. A particle is projected with a speed, $v=\sqrt{2}{\text{ms}}^{-1}$ and at ${45}^{\circ }$ with the horizontal as shown in the figure. At the very instant of projection, the cable of the elevator breaks, and the elevator starts falling freely. What will be the separation between the particle and its image (in $\text{m}$), $0.5\text{s}$ after the instant of projection?

Answer 1

Given:

$v=\sqrt{2}{\text{ms}}^{-1}$

$\theta ={45}^{\circ }$ with horizontal

$\therefore v_x=v\cos {45}^{\circ }=1{\text{ms}}^{-1}$

$v_y=v\sin {45}^{\circ }=1{\text{ms}}^{-1}$

We will work in the elevator frame of reference.

In vertical direction:

Relative acceleration,

$a_r=a_p-a_e=g-g=0$

Relative initial velocity,

$v_r=v_p-v_e=1-0=1{\text{ms}}^{-1}$

Using kinematic equation of motion for relative motion,

$S_r=v_{r}t+\frac{1}{2}{a}_r{t}^2$

$\Rightarrow S_r=1\times 0.5=0.5\text{m}$

Hence, separation between the particle and its image will be,

$d=2\times S_r=2\times 0.5=1\text{m}$