An athlete runs up and projects Javelin at a certain angle to the horizontal. If the speed of the Javelin relative to the athlete is equal to the speed of the athlete at the time of projection, what should be the angle at which he should throw relative to himself so that the range is maximum? The height of the athlete can be considered negligible.

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$45^{0}$

$30^{0}$

$60^{0}$

$37^{0}$

answer is C.

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Detailed Solution

$\begin{array}{l} (u + u \cos θ) = u_{x}, u \sin θ = u_{y} \\ R = \frac{2 v^{2}}{g} (\cos θ + 1) \sin θ, \frac{d R}{d θ} = 0 \\ (\cos θ) (\cos θ + 1) + (- \sin θ) \cdot \sin θ = 0 \\ 2 \cos^{2} θ + \cos θ - 1 = 0 \\ on solving \cos θ \Rightarrow + \frac{1}{2} \Rightarrow θ = 60^{\circ} \end{array}$

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