A cannon on a level plain is aimed at an angle α above the horizontal and a shell is fired with a muzzle velocity vo towards a vertical cliff a distance R away. Then the height from the bottom at which the shell strikes the side walls of the cliff is

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R sin⁡ α−12gR2v02sin2⁡α

R cos ⁡α−12gR2v02cos2⁡α

R tan ⁡α−12gR2v02cos2⁡α

R tan⁡ α−12gR2v02sin2⁡α

answer is C.

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Detailed Solution

The time taken to move a horizontal distance R is t = R/(v0 cos α). Therefore, the vertical distance moved in this time is given by h=uyt−12gt2=v0sin⁡α×Rv0cos⁡α−12gRv0cos⁡α2=R tan⁡ α−gR22v02cos2⁡α

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