Question

If R is the range of a projectile on a horizontal plane and h is the maximum height, then maximum horizontal range with the same velocity of projection is

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

Answer 1

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

$\text{Step 1 : Range of the projectile}$

Let u be the velocity of projection and $\theta$ be the angle of projection

$R=\frac{u^2\sin 2\theta }{g}$ $....\left(1\right)$

Range is maximum when, $\sin 2\theta =1$

$\Rightarrow R_{max}=u^2/g$ $....\left(2\right)$

$\text{Step 2 : Maximum height attained}$

$h=\frac{u^2{\sin }^2\theta }{2g}$ $....\left(3\right)$

$\text{Step 3 : Solving above equations}$

Dividing equation $\left(3\right)$ by equation $\left(1\right)$, We get:

$\frac{h}{R}=\frac{\tan \theta }{4}$

$\Rightarrow \tan \theta =\frac{4h}{R}$

$\Rightarrow \sin \theta =\frac{4h}{\sqrt{R^2+16h^2}}$ $....\left(4\right)$

$\text{Step 4 : Maximum range}$

From equation (2)

$R_{max}=u^2/g$

Putting value of $u^2/g$ from equation (3) in above equation, We get:

$R_{max}=\frac{2h}{{\sin }^2\theta }$

Putting the value of ${\sin }^2\theta$ from equation $\left(4\right)$, We get:

$R_{max}=\frac{2h(R^2+16h^2)}{16h^2}$

$\Rightarrow R_{max}=\frac{R^2}{8h}+2h$

Hence, correct option is $\left(D\right)$