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Questions  

A cannon on a level plane is aimed at an angle θ above the horizontal and a shell is fired with a muzzle velocity v0 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
Dsin⁡θ−gD22v02sin2⁡θ
b
Dcos⁡θ−gD22v02cos2⁡θ
c
Dtan⁡θ−gD22v02cos2⁡θ
d
Dtan⁡θ−gD22v02sin2⁡θ

detailed solution

Correct option is C

The time taken to move a horizontal distance D is given byt=D/v0cos⁡θThe vertical distance moved in this timeh=v0sin⁡θ×D/v0cos⁡θ−12gD/v0cos⁡θ2=Dsin⁡θcos⁡θ−gD22v02cos2⁡θ=Dtan⁡θ−gD22v02cos2⁡θ

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Reason : Range of projectile does not depend on the initial velocity of a body.


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