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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a
R sin α−12gR2v02sin2α
b
R cos α−12gR2v02cos2α
c
R tan α−12gR2v02cos2α
d
R tan α−12gR2v02sin2α
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=v0sinα×Rv0cosα−12gRv0cosα2=R tan α−gR22v02cos2α