Question

If $R$ is the horizontal range for an inclination and $h$ is the maximum height reached by the projectile, then the maximum range is given by-

• A. $\frac{R^2}{8h}-2h$
• B. $\frac{R^2}{8h}+2gh$
• C. $\frac{R^2}{8h}+2h$
• D. $\frac{R^2}{8h}$

Answer 1

The correct option is C $\frac{R^2}{8h}+2h$

Let the projectile is projected with a velocity $u$ making an angle $\theta$ with the horizontal.

$\therefore$ Maximum height reached by the projectile is given by $H_{max}=h=\frac{u^{2}si{n}^2\theta }{2g}$ .......(1)

Also horizontal range is given by $R=\frac{u^{2}sin2\theta }{g}$ ...............(2)

Dividing equation (1) and (2) we get, $\frac{h}{R}=\frac{si{n}^2\theta }{2sin2\theta }$

$\therefore$ $\frac{h}{R}=\frac{si{n}^2\theta }{2\times 2sin\theta cos\theta }$ $⟹\frac{4h}{R}=tan\theta$ ..........(3)

Equation (3) implies $sin\theta =\frac{4h}{\sqrt{16h^2+R^2}}$

Now maximum range of the projectile $R_{max}=\frac{u^2}{g}$ .......(4)

From (1) and (4), $h=\frac{R_{max}\times 16h^2}{2(16h^2+R^2)}$

$⟹R_{max}=2h+\frac{R^2}{8h}$