Q. A car is travelling at a velocity of $10\ kmh^{- 1}$ on a straight road. The driver of the car throws a parcel with a velocity of $10\sqrt{2}\ kmh^{- 1}$ when the car is passing by a man standing on the side of the road. If the parcel is to reach the man, the direction of throw makes the following angle with direction of the car
A) $135{^\circ}$
B) $45{^\circ}$
C) $\tan^{- 1}{(\sqrt{2})60{^\circ}}$
D) $\tan\left( \frac{1}{\sqrt{2}} \right)$
Explanation
(b) Let $v_{1}$ be the velocity of the car and $v_{2}$ be the velocity of the parcel. The parcel is thrown at an angle $\theta$ from $O$. It reaches the mass at $M$.
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$\therefore\ \cos{\theta = \frac{v_{1}\ \ \ }{v_{2}} = \frac{10}{10\sqrt{2}}} = \frac{1}{\sqrt{2}} = \cos{45{^\circ}}$
So $\theta = 45{^\circ}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.