Q.
A man standing on a hill top projects a stone horizontally with speed $v_{0}$ as shown in figure. Taking the coordinate system as given in the figure. The coordinates of the point where the stone will hit the hill surface
A) $\left( \frac{2v_{0}^{2}tan\theta}{g},\frac{- 2v_{0}^{2}\tan^{2}\theta}{g} \right)$
B) $\left( \frac{2v_{0}^{2}}{g},\frac{2v_{0}^{2}\tan^{2}\theta}{g} \right)$
C) $\left( \frac{2v_{0}^{2}\tan\theta}{g},\frac{2v_{0}^{2}}{g} \right)$
D) $\left( \frac{2v_{0}^{2}\tan^{2}\theta}{g},\frac{2v_{0}^{2}\tan\theta}{g} \right)$
Explanation
(a) Range of the projectile on an inclined plane (down the plane) is,
$R = \frac{u^{2}}{gcos^{2}\beta}\lbrack\sin(2\alpha + \beta) + \sin\beta\rbrack$
Here, $u = v_{0},\alpha = 0$ and $\beta = \theta$
$\therefore R = \frac{2v_{0}^{2}\sin\theta}{g\cos^{2}\theta}$
[image:4_image121.jpeg]
Now $x = R\cos\theta = \frac{2v_{0}^{2}\tan\theta}{g}\ $
and $y = - R\sin\theta = \ - \frac{2v_{0}^{2}\tan^{2}\theta}{g}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.