Question

A particle $ P $ falling vertically from a height hits a plane inclined to the horizontal at an angle $ \theta $ with speed $ v $ and rebounds elastically, as shown. The distance along the plane where it hits the second time is

• A. $ \frac{4v^{2}\,\sin^{2}\,\theta}{g} $
• B. $ \frac{v^{2}\,\sin^{2}\,\theta}{2g} $
• C. $ \frac{4v^{2}\,\sin\,\theta}{g} $
• D. $ \frac{v^{2}\,\sin\,\theta}{4g} $

Answer 1

Answer: (C)

We have two components for each initial velocity $(v_{0})$ and acceleration due to gravity $(g)$. In this frame of reference the horizontal and vertical component of velocity and gravity are given by
$V_{x}=V_{0} \sin\, \theta, v_{y}=V_{0}\,\cos\,\theta,$
$g_{x}=g\,\sin\,\theta, g_{y}=g\,\cos\, \theta$
Now, $t_{bounce}=\frac{2V_{y}}{g_{y}}=\frac{2V_{0}\,\cos\,\theta}{g\,\cos\,\theta}=\frac{2V_{0}}{g}$
For horizontal distance $d=v_{x}t+\frac{1}{2}g_{x}t^{2}$
$\therefore d=v_{0}\, \sin \,\theta \left(\frac{2V_{0}}{g}\right)+\frac{1}{2} g \sin\,\theta \left[\frac{2V_{0}}{g}\right]^{2}$
$\frac{=2v_{0}^{2}\sin\,\theta}{g}+\frac{1}{2}g \frac{4V^{2}}{g^{2}}\sin\,\theta=\frac{4V_{0}^{2} \sin\,\theta}{g}$