Question

A car is moving horizontally along a straight line with constant speed $30m/s$ . A projectile is to be fired from the moving car in such a way that it will return to the car after the car has moved through $60\sqrt{3}m$. The speed and angle at which the projectile must be projected are respectively:

• A. $20\sqrt{3}m/s;{45}^0$
• B. $20m/s;{30}^0$
• C. $20\sqrt{3}m/s;{30}^0$
• D. $20\sqrt{3}m/s;{60}^0$

Answer 1

Answer: (C)

$20\sqrt{3}m/s;{30}^0$

Range $=60\sqrt{3}=\frac{u^2\sin 2\theta }{g}$
$u\cos \theta =30-\left(1\right)$
$\Rightarrow 60\sqrt{3}=\frac{2u^2\sin \theta \cos \theta }{g}$
$\Rightarrow 30\sqrt{3}=\frac{\left(2u\sin \theta \right)\times 30}{g}$
$\Rightarrow 10\sqrt{3}=2u\sin \theta$
$\Rightarrow u\sin \theta =5\sqrt{3}-\left(2\right)$
Solve (1) & (2) we get
$u=20\sqrt{3};\theta ={30}^0$