Two particles are projected upwards with the same initial velocity v₀ in two different angles of projection such that their horizontal ranges are the same. The ratio of the heights of their highest points will be

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$\tan^{2} θ_{1}$

$v_{0} \sin θ_{1}$

$v_{0} / \cos θ_{1}$

${v_{0}}^{2} \sin^{2} θ_{1}$

answer is A.

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Detailed Solution

As the horizontal ranges are the same

$\begin{array}{l} \frac{v_{0}^{2} \sin 2 θ_{1}}{g} = \frac{v_{0}^{2} \sin 2 θ_{2}}{g} \\ or \sin 2 θ_{1} = \sin 2 θ_{2} \\ or 2 θ_{1} = π - 2 θ_{2} or θ_{1} + θ_{2} = π / 2 \\ Now {(h_{1})}_{max} = \frac{{(v_{0} \sin θ_{1})}^{2}}{2 g} \\ and {(h_{2})}_{max} = \frac{{(v_{0} \sin θ_{2})}^{2}}{2 g} \\ ∴ \frac{{(h_{1})}_{max}}{{(h_{2})}_{max}} = \frac{{(\sin θ_{1})}^{2}}{{(\sin θ_{2})}^{2}} = \frac{\sin^{2} θ_{1}}{\cos^{2} θ_{1}} = \tan^{2} θ_{1} \end{array}$