Question

An elevator at rest which is at ${10}^{th}$ floor of a building is having a plane mirror fixed on its floor. A particle is projected with speed $\sqrt{2}m{s}^{-1}$ and at $\angle {45}^o$ with horizontal. At the very instant of projection, the cable of elevator breaks and starts falling freely. What will be separation between particle and image, $0.5sec$ after instant of projection.

Answer 1

For particle:
$\begin{array}{c}h=1\times 0.5-\frac{1}{2}\times 10\times \frac{1}{4}\\ =\frac{1}{2}-\frac{5}{4}=-\frac{3}{4}\end{array}$
for elevator
$h_1=-\frac{1}{2}\times 10\times \frac{1}{4}=-\frac{5}{4}$
then distance between particle and mirror
$=\frac{5}{4}-\left(-\frac{3}{4}\right)=-\frac{1}{4}$
then if
$\begin{array}{c}u=\frac{1}{4}\\ =\frac{1}{4}+\frac{1}{4}=\frac{2}{4}=\frac{1}{2}m\end{array}$