A ball is projected from the ground with a speed $15 {ms}^{- 1}$ at an angle $θ$ with horizontal so that its range and maximum height are equal, then $' \tan θ'$ will be equal to

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$\frac{1}{2}$

$\frac{1}{4}$

answer is D.

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Detailed Solution

$\begin{array}{l} R = H \\ \frac{2 v_{x} \times v_{y}}{g} = \frac{v_{y}^{2}}{2 g} \\ v_{x} = \frac{v_{y}}{4}; ucos θ = \frac{usin θ}{4} \\ \tan θ = 4 \end{array}$

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