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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tan−1(θ/t)

tan−1(v cos θv sinθ−gt)

tan−1(v sinθ−gtv cos θ)

answer is D.

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Detailed Solution

υ'cosα=υcosθAnd υ'sinα=υsinθ−gt∴ υ'sinαυ'cosα=υsinθ−gtυcosθor tanα=υsinθ−gtυcosθ∴ α=tan−1(υsinθ−gtυcosθ)=tan−1(vsinθ−gtvcosθ)

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