A projectile is fired at an angle of 45∘ with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection is
A
tan−1(√32)
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B
45∘
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C
60∘
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D
tan−112
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Solution
The correct option is Dtan−112
Let α be the elevation angle ⇒tanα=HR/2.......(i) By using the relation, 4H=Rtanθ HR=tanθ4......... (ii) Using (i) and (ii), we get tanα=tanθ2 ⇒tanα=tan45∘2 α=tan−1(12)