Question bank/JEE Main/Physics

The speed of a projectile at the highest point is $v_{1}$ and at
the point half the maximum height is $v_{2}$. If
$\frac{v_{1}}{v_{2}} = \sqrt{\frac{2}{5}}$, then the angle of projection
is π/n. Find n.

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$0 = {(v_{2}sin\beta)}^{2} - 2g\frac{H}{2}$
$v_{1}^{2} = v_{1}^{2} - gh$
Also $\frac{v_{1}}{v_{2}} = \sqrt{\frac{2}{5}}$
From above
$v_{1} = \sqrt{\frac{2gH}{3}}$
$v_{2} = \sqrt{\frac{5gH}{3}}$
$H = \frac{u^{2}\sin^{2}\theta}{2g} \Rightarrow sin2\theta = \frac{2gH\cos^{2}\theta}{v_{1}^{2}}$
$\tan^{2}\theta = \frac{2gH}{v_{1}^{2}}{or\ tan}^{2}\theta = \frac{2gH \times 3}{2gH}$
$tan\theta = \sqrt{3}$
$\theta = 60^{\circ}$

Source: Grass-G / P-18 Physics paper 27 Nov Third.docx