If the time of flight of a bullet over a horizontal range R is T, then inclination of the direction of projection with the horizontal is
A
tan−1[gT22R]
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B
tan−1[2R2gT]
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C
tan−1[2Rg2π]
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D
tan−1[2RgT]
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Solution
The correct option is Atan−1[gT22R] Time of flight, T=2vsinθg T2=4v2sin2θg2......(i) We know that, Range,R=v2(2sinθcosθ)g......(ii) Dividing (i) from (ii), we get T2R=2gtanθ tanθ=gT22R ⇒θ=tan−1[gT22R]