Question

The maximum range of a bullet fired from a toy pistol mounted on a car at rest ${R_0} = 40m$. 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 $v = \dfrac{{20m}}{s}$ on a horizontal surface? ($g = \dfrac{{10m}}{{{s^2}}}$)
(A) $30^\circ$
(B) $60^\circ$
(C) $75^\circ$
(D) $45^\circ$

Answer 1

Solve this question from two frames of references. Firstly, from the car as frame of reference calculate the velocity of the projectile (gun) and then from ground as a frame of reference, calculate the condition of the maximum range- that will give us the value of the projectile angle.

Complete Step-by-step Solution:
When the car is at rest: The piston is shot with initial velocity $u$. In calculating range, the horizontal component of velocity will come to play i.e. $u\cos \theta$ $u\cos \theta$
The range of a projectile is given by-
$R = \dfrac{{2u\sin \theta u\cos \theta }}{g}$
$R = \dfrac{{{u^2}\sin 2\theta }}{g}$
For range to be maximum –
$\sin 2\theta = 1$
$\Rightarrow 2\theta = \dfrac{\pi }{2}$
$\Rightarrow \theta = \dfrac{\pi }{4}$
We have the maximum range:
${R_{\max }} = {R_0} = \dfrac{{{u^2}}}{g}$
$\Rightarrow 40 = \dfrac{{{u^2}}}{{10}}$
$\Rightarrow {u^2} = 400$
$\Rightarrow u = \dfrac{{20m}}{s}$
When the car is moving: we will take ground as a frame of reference.
In ground frame:
$\Rightarrow R = \dfrac{{2{u_x}{u_y}}}{g}$
$\Rightarrow R = \dfrac{{2(20 + u\cos \theta )u\sin \theta }}{g}$
For the range to be maximum, its derivative with respect to angle will be zero.
$\Rightarrow \dfrac{{dR}}{{d\theta }} = 0 \\\ \Rightarrow R = \dfrac{{2(20u\sin \theta + {u^2}\sin \theta \cos \theta )}}{g} \\\ \Rightarrow \dfrac{{dR}}{{d\theta }} = \dfrac{{2(20u\cos \theta + {u^2}({{\cos }^2}\theta - {{\sin }^2}\theta )}}{g} \\\ \Rightarrow \dfrac{{dR}}{{d\theta }} = \dfrac{{2(20u\cos \theta + {u^2}(2{{\cos }^2}\theta - 1))}}{g} \\\ \Rightarrow \dfrac{{dR}}{{d\theta }} = 0 \\\ \Rightarrow 20\cos \theta + u(2{\cos ^2}\theta - 1) = 0 \\\ \Rightarrow 20\cos \theta + 20(2{\cos ^2}\theta - 1) = 0 \\\ \Rightarrow \cos \theta + 2{\cos ^2}\theta - 1 = 0 \\\ \Rightarrow \cos \theta = - 1,\dfrac{1}{2} \\\ \Rightarrow \cos \theta = \dfrac{1}{2} \\\ \Rightarrow \theta = 60^\circ \\\$
The required inclination is $60^\circ$.

Hence, the correct option is B.

Note- In determining the maximum range, the horizontal component is used. In determining the vertical height, the vertical component of velocity is used.