Question

If $R$ and $H$ are the horizontal range and maximum height attained by a projectile, than its speed of projection is

• A. $\sqrt{2gR+\frac{4R^2}{gH}}$
• B. $\sqrt{2gH+\frac{R^{2}g}{8H}}$
• C. $\sqrt{2gH+\frac{8H}{Rg}}$
• D. $\sqrt{2gH+\frac{R^2}{H}}$

Answer 1

Answer: (B)

$H=\frac{u^2{\sin }^2\theta }{2g}\Rightarrow \sin \theta =\sqrt{\frac{2gH}{u^2}}$
$R=\frac{2u^2\sin \theta \cos \theta }{g}$
$R=\frac{2u^2}{g}\sqrt{\frac{2gH}{u^2}}\times \sqrt{1-\frac{2gH}{u^2}}$
$R=\frac{2u^2}{g}\sqrt{\frac{2gH}{u^2}}\times \sqrt{\frac{u^2-2gH}{u^2}}$
$\frac{gR}{2\sqrt{2gH}}=\sqrt{u^2-2gH}$
Squaring both the sides,
$\frac{g{R}^2}{4\times 2gH}=u^2-2gH$
$\Rightarrow u^2=2gH+\frac{9R^2}{8H}$
$u=\sqrt{2gH+\frac{g{R}^2}{8H}}$