Question

The maximum range of a bullet fired from a toy pistol mounted on a car at rest is $R_0=40\text{m}$. What will be the acute angle of inclination of the pistol for maximum range when the car is moving in the direction of firing with uniform velocity $v=20\text{m/s}$ on a horizontal surface ?
$($ Take $g=10{\text{m/s}}^2)$

• A. ${60}^{\circ }$
• B. ${30}^{\circ }$
• C. ${75}^{\circ }$
• D. ${45}^{\circ }$

Answer 1

The correct option is A ${60}^{\circ }$

Given:
$R_0=40\text{m};v=20\text{m/s}$

Let $u$ be the projected speed of the bullet.

As we know that,

$R_{max}=R_0=\frac{u^2}{g}$

$\Rightarrow 40=\frac{u^2}{10}$

$\therefore u=20\text{m/s}$


Let $\theta$ be the inclination of the piston for maximum range when car is moving.

$\therefore R=\frac{2u_x{u}_y}{g}=\frac{2(v+u\cos \theta )u\sin \theta }{g}$

For maximum range, $\frac{d{R}_{max}}{d\theta }=0$

$\frac{2u}{g}\left[\right(-u\sin \theta )\sin \theta +(v+u\cos \theta \left)\cos \theta \right]=0$

$-u{\sin }^2\theta +v\cos \theta +u{\cos }^2\theta =0$

$-(1-{\cos }^2\theta )+\cos \theta +{\cos }^2\theta =0(\because u=v=10\text{m/s})$

$2{\cos }^2\theta +\cos \theta -1=0$

$\cos \theta =\frac{-1\pm \sqrt{1-4\left(2\right)(-1)}}{2\left(2\right)}=-1,\frac{1}{2}$

For acute angle,

$\cos \theta =\frac{1}{2}\Rightarrow \theta ={60}^{\circ }$

Hence, option $\left(\text{A}\right)$ is correct.