Question:
The maximum range of a bullet fired from a toy pistol mounted on a car at rest is \(R_0=40\) 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 \(v=20\) m/s on a horizontal surface? (\(g=10\ \mathrm{m/s^2}\))
The maximum range of a bullet fired from a toy pistol mounted on a car at rest is \(R_0=40\) 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 \(v=20\) m/s on a horizontal surface? (\(g=10\ \mathrm{m/s^2}\))
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Find the muzzle speed from \(R_0=u^2/g\); the bullet's ground velocity is the vector sum with the car velocity; maximise \(R\).
Updated On: Nov 4, 2026
- \(30^\circ\)
- \(60^\circ\)
- \(75^\circ\)
- \(45^\circ\)
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The Correct Option is B
Solution and Explanation
At rest: \(R_0=\frac{u^2}{g}=40\Rightarrow u^2=400,\ u=20\) m/s (relative to the car).
With the car moving, the horizontal velocity of the bullet is \(u\cos\theta+v\) and the vertical is \(u\sin\theta\). Time of flight \(T=\frac{2u\sin\theta}{g}\).
Range \(R=(u\cos\theta+v)\frac{2u\sin\theta}{g}\propto\sin\theta(1+\cos\theta)\) (since \(v=u=20\)).
\(\frac{dR}{d\theta}=0:\ \cos\theta(1+\cos\theta)-\sin^2\theta=0\Rightarrow2\cos^2\theta+\cos\theta-1=0\Rightarrow\cos\theta=\frac12\).
\(\theta=60^\circ\).
Answer: (2) \(60^\circ\)
With the car moving, the horizontal velocity of the bullet is \(u\cos\theta+v\) and the vertical is \(u\sin\theta\). Time of flight \(T=\frac{2u\sin\theta}{g}\).
Range \(R=(u\cos\theta+v)\frac{2u\sin\theta}{g}\propto\sin\theta(1+\cos\theta)\) (since \(v=u=20\)).
\(\frac{dR}{d\theta}=0:\ \cos\theta(1+\cos\theta)-\sin^2\theta=0\Rightarrow2\cos^2\theta+\cos\theta-1=0\Rightarrow\cos\theta=\frac12\).
\(\theta=60^\circ\).
Answer: (2) \(60^\circ\)
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