Question

If $R$ and $H$ are the horizontal range and maximum height attained by a projectile , then 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

The correct option is B $\sqrt{2gh+\frac{R^{2}g}{8H}}$

$\text{Step 1: Motion in y direction [Refer Fig.]}$

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

The maximum height of the projectile is given by:

$H=\frac{u^{2}si{n}^2\theta }{2g}$

$\Rightarrow u\sin \theta =\sqrt{2gH}$ $....\left(1\right)$

$\text{Step 2: Calculate time of flight(T)}$

From B to C in y direction

$u_y=0,$ $t=\frac{T}{2}$, $s_y=-H$; $a_y=-g$ , where, $T=$ Total time of flight

Apply second equation of motion

$s=ut+\frac{1}{2}a{t}^2$

$\Rightarrow$ $-H=0-\frac{1}{2}g{\left(\frac{T}{2}\right)}^2$

$\Rightarrow$ $T=\sqrt{\frac{8H}{g}}$

$\text{Step 3: Motion in x direction}$

Since acceleration in $x$ direction is zero

$R=u\cos \theta \times T$

$\Rightarrow R=u\cos \theta \sqrt{\frac{8H}{g}}$

$\Rightarrow u\cos \theta =R\sqrt{\frac{g}{8H}}$ $....\left(2\right)$

$\text{Step 4: Calculate velocity of projection}$

Squaring and adding equation $\left(1\right)$ and $\left(2\right)$, we get

$u^2=2gH+\frac{R^{2}g}{8H}$

$\Rightarrow u=\sqrt{2gH+\frac{R^{2}g}{8H}}$

Hence option $B$ is correct.