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A body is projected horizontally with velocity $19.6 m s^{- 1}$ from certain height. After 2s, the angle made by its direction of motion to the horizontal is

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$90^{0}$

$60^{0}$

$45^{0}$

$0^{0}$

answer is C.

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Detailed Solution

$u = 19.6 m, t = 2 s h = \frac{t^{2} g}{2} h = 19.6 m T a n θ = \frac{\sqrt{2 g h}}{u} \Rightarrow θ = 45^{0}$

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