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Q.
If $R$ and $H$ are the horizontal range and maximum height attained by a projectile, than its speed of projection is
Motion in a Plane
Solution:
$H=\frac{u^{2} \sin ^{2} \theta}{2 g} \Rightarrow \sin \theta=\sqrt{\frac{2 g H}{u^{2}}}$
$R=\frac{2 u^{2} \sin \theta \cos \theta}{g}$
$R=\frac{2 u^{2}}{g} \sqrt{\frac{2 g H}{u^{2}}} \times \sqrt{1-\frac{2 g H}{u^{2}}}$
$R=\frac{2 u^{2}}{g} \sqrt{\frac{2 g H}{u^{2}}} \times \sqrt{\frac{u^{2}-2 g H}{u^{2}}}$
$\frac{g R}{2 \sqrt{2 g H}}=\sqrt{u^{2}-2 g H}$
Squaring both the sides,
$\frac{g R^{2}}{4 \times 2 g H}=u^{2}-2 g H$
$\Rightarrow \quad u^{2}=2 g H+\frac{9 R^{2}}{8 H}$
$u=\sqrt{2 g H+\frac{g R^{2}}{8 H}}$