For a projectile projected from ground at an angle θ with horizontal, gT2=2R√3, where T is time of fight, R is horizontal range of projectile, g is acceleration due to gravity. The angle of projection (θ) is
A
30∘
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B
60∘
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C
45∘
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D
90∘
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Solution
The correct option is A30∘ According to question: gT2=2R√3⇒g(2usinθg)2=2√3u2sin2θg g(4u2sin2θg2)=2√3×u2sin2θg 4u2sin2θg=2√3×u2×2sinθcosθg
On solving we get tanθ=1√3⇒θ=30∘