A cannon on a level plane is aimed at an angle $θ$ above the horizontal and a shell is fired with a muzzle velocity $v_{o}$ 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

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D cos

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D tan

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D tan

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Solution

The correct option is C D tan -
Equation of trajectory for oblique projectile motion

y = $x t a n θ - \frac{g x^{2}}{2 u^{2} c o s^{2} θ}$

Substituting x = D and u = $v_{0}$
h = $D t a n θ - \frac{g D^{2}}{2 u_{0}^{2} c o s^{2} θ}$

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

The correct option is C D tan - Equation of trajectory for oblique projectile motion y = x tan θ−gx22u2cos2θ Substituting x = D and u = v0 h = D tan θ−gD22u20 cos2θ

A cannon on a level plane is aimed at an angle $θ$ above the horizontal and a shell is fired with a muzzle velocity $v_{o}$ 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

The correct option is C D tan -
Equation of trajectory for oblique projectile motion

y = $x t a n θ - \frac{g x^{2}}{2 u^{2} c o s^{2} θ}$

Substituting x = D and u = $v_{0}$
h = $D t a n θ - \frac{g D^{2}}{2 u_{0}^{2} c o s^{2} θ}$