The angle which the velocity vector of a projectile thrown with a velocity v at an angle θ to the horizontal will make with the horizontal after time t of its being thrown up is
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a
θ
b
tan−1(θ/t)
c
tan−1VcosθVsinθ−gt
d
tan−1vsinθ−gtvcosθ
answer is D.
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Detailed Solution
See fig. (5). Here v′cosα=vcosθand v′sinα=vsinθ−gt∴ V′sinαV′cosα=Vsinθ−gtVcosθ or tanα=Vsinθ−gtVcosθα=tan−1(Vsinθ−gt)Vcosθ