Q. A cannon on a level plane is aimed at an angle $\theta$ above the horizontal and a shell is fired with a muzzle velocity $v_{0}$ towards a vertical cliff a distance $D$ away. Then the height from the bottom at which the shell strikes the side walls of the cliff is
A) $D\sin\theta - \frac{gD^{2}}{2v_{0}^{2}\sin^{2}\theta}$
B) $D\cos\theta - \frac{gD^{2}}{2v_{0}^{2}\cos^{2}\theta}$
C) $D\tan\theta - \frac{gD^{2}}{2v_{0}^{2}\cos^{2}\theta}$
D) $D\tan\theta - \frac{gD^{2}}{2v_{0}^{2}\sin^{2}\theta}$
Explanation
(c) Equation of trajectory for oblique projectile motion
$y = x\tan\theta - \frac{gx^{2}}{2u^{2}\cos^{2}\theta}$
Substituting $x = D$ and $u = v_{0}$
$h = D\tan\theta - \frac{gD^{2}}{2u_{0}^{2}\cos^{2}\theta}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.