Question

If the range and maximum height of a projectile are respectively $R$ and $H$, the maximum range that could be obtained with the same velocity of projection is

• A. $4H$
• B. $2R$
• C. $2H+\frac{R^2}{8H}$
• D. $2R+\frac{H^2}{8R}$

Answer 1

Answer: (C)

$2H+\frac{R^2}{8H}$

We know that: $H=\frac{u^2{\sin }^2\theta }{2g}$

$⟹u^2{\sin }^2\theta =2gH$ -----$\left(1\right)$

$R=\frac{u^2\sin 2\theta }{g}$

$R=\frac{2u^2\sin \theta \cos \theta }{g}$

$R^2=\frac{4u^{4}si{n}^2\theta co{s}^2\theta }{g^2}$

$u^{2}co{s}^2\theta =\frac{g^2{R}^2}{4u^{2}si{n}^2\theta }$

$u^2{\cos }^2\theta =\frac{g{R}^2}{8H}$ ------$\left(2\right)$

Adding $\left(1\right)and\left(2\right)$: $u^2=\frac{g{R}^2}{8H}+2gH$

Maximum range for a given projectile is $\frac{u^2}{g}$.
Hence, maximum range is $=2H+\frac{R^2}{8H}$