Question

A cannon on a level plane is aimed at an angle θ 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 θ - $\frac{gD^{2}}{2{v_{0}}^{2} \sin^{2} \theta}$
(b) D cos θ - $\frac{gD^{2}}{2{v_{0}}^{2} \cos^{2} \theta}$
(c) D tan θ - $\frac{gD^{2}}{2{v_{0}}^{2} \cos^{2} \theta}$
(d) D tan θ - $\frac{gD^{2}}{2{v_{0}}^{2} \sin^{2} \theta}$

Answer 1

A cannon on a level plane is aimed at an angle of theta( θ) above the horizontal and the shell is fired with the muzzle velocity v towards the vertical cliff at a distance D away . Thus,

The equation of trajectory for oblique projectile motion is -

y = x tan  θ- gx²/ 2u²cos² θ

Here, x = D and u = v0

Therefore, h = D tan θ - gD²/ 2v²cos² θ

Therefore the he height from the bottom at which the shell strikes the side walls of cliff is Dtan θ - gR²/ 2v²cos² θ