Q. A cannon of a level plane is aimed at an angle $\theta$ above the horizontal and a shell is fired muzzle velocity $v_{0}$ towards a cliff $D$ distance away. The height at which the canon strikes the cliff is given by
A) $D\sin\theta - \frac{1}{2}\frac{gD^{2}}{v_{0}^{2}\sin^{2}\theta}\ $
B) $D\cos\theta - \frac{1}{2}\frac{gD^{2}}{v_{0}^{2}\sin^{2}\theta}$
C) $D\tan\theta - \frac{1}{2}\frac{gD^{2}}{v_{0}^{2}\cos^{2}\theta}$
D) $D\tan\theta - \frac{1}{2}\frac{gD^{2}}{v_{0}^{2}\sin^{2}\theta}$
Explanation
(c) Time taken to cover horizontal distance $D$ with constant horizontal velocity, $t = D/v_{0}\cos\theta$. Taking vertical motion for time $t$, we have
$h = v_{0}\sin\theta \times t - \frac{1}{2}gt^{2}$
$= v_{0}\sin\theta \times \frac{D}{v_{0}\cos\theta} - \frac{1}{2}g\left( \frac{D}{v_{0}\cos\theta} \right)^{2}$
$= D\tan\theta - \frac{1}{2}\frac{gD^{2}}{v_{0}^{2}\cos^{2}\theta}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.