Question

A particle is projected with speed $25m/s$ at angle ${53}^o$ from horizontal in front of an inclined plane mirror as shown in figure. Choose the correct options: ($g=10m/s^2$)

• A. Speed of image w.r.t object will be minimum after time $\frac{7}{8}sec$ of projection
• B. Speed of image w.r.t object will be maximum after time $4$ sec of projection
• C. Minimum speed of image w.r.t object is zero
• D. Maximum speed of image w.r.t object is $50m/s$

Answer 1

The correct option is C Minimum speed of image w.r.t object is zero

Horizontal velocity $v_H=25\times \mathrm{Cos}{53}^{\circ }\text{m}\text{.}{\text{s}}^{-1}\approx 15\text{m}\text{.}{\text{s}}^{-1}$

Vertical velocity at $t=\frac{7}{8}\sec$, $v_V=\left(25\times \mathrm{Sin}{53}^{\circ }\right)-g\times \left(\frac{7}{8}\right)$

$\therefore v_V=\left(25\times \frac{4}{5}\right)-10\times \left(\frac{7}{8}\right)\approx \frac{45}{4}\text{m}\text{.}{\text{s}}^{-1}$

Resultant velocity of the image at $t=\frac{7}{8}\sec$-

$v=\left(v_H\times \mathrm{Sin}{37}^{\circ }\right)-\left(v_V\times \mathrm{Cos}{37}^{\circ }\right)$

$v=\left(15\times \frac{3}{5}\right)-\left(\frac{45}{4}\times \frac{4}{5}\right)=9-9=0$

It is the minimum velocity of the image.