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2015-10-02T12:32:42+05:30

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Horizontal range, R= \frac{u^2sin2\theta}{g}

maximum height, H= \frac{u^2sin^2\theta}{2g}

Given that R = 4H

\Rightarrow \frac{u^2sin2\theta}{g} =4 \times \frac{u^2sin^2\theta}{2g} \\ \\ \Rightarrow \frac{2  u^2sin\theta cos \theta }{g} =4 \times \frac{u^2sin^2\theta}{2g}\\ \\ \Rightarrow \frac{2  u^2sin\theta cos \theta }{g} = \frac{2u^2sin^2\theta}{g}\\ \\ \Rightarrow cos \theta=sin\theta\\ \\ \Rightarrow  \frac{sin\theta}{cos \theta}=tan \theta=1\\ \\ \Rightarrow \theta = 45^o

Angle of projection is 45°
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