A cannon on a level plane is aimed at an angle \emptyset ab…

Physics Motion in a Plane Mix Single Correct MCQ

Published on: September 12, 2026

A cannon on a level plane is aimed at an angle \emptyset above the horizontal and a shell is fired with a muzzle velocity v_o 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

\(D \sin \theta - \frac{g D^{2}}{2 v_{0}^{2} \sin^{2} \theta}\)

\(D \cos \theta - \frac{g D^{2}}{2 v_{0}^{2} \cos^{2} \theta}\)

\(D \tan \theta - \frac{g D^{2}}{2 v_{0}^{2} \cos^{2} \theta}\)

\(D \tan \theta - \frac{g D^{2}}{2 v_{0}^{2} \sin^{2} \theta}\)

Text Solution

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The correct answer is:

Equation of trajectory for oblique projectile motion

\(y = x \tan \theta - \frac{g x^{2}}{2 u^{2} \cos^{2} \theta}\)

Substituting X = D and u = V_0

\(h = D \tan \theta - \frac{g D^{2}}{2 u_{0}^{2} \cos^{2} \theta}.\)

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