Q. A particle $A$ is projected from the ground with an initial velocity of $10\ ms^{- 1}$ at an angle $60{^\circ}$ with horizontal. From what height should an another particle $B$ be projected horizontally with velocity $5\ ms^{- 1}$ so that both the particles collide in ground at point $C$ if both are projected simultaneously $g = 10ms^{- 2}$
A) 10 m
B) 15 m
C) 20 m
D) 30 m
Explanation
(b) Both the particles will meet at $C$, if their time of flight is the same. The time of flight of $A\ $is
$T = \frac{2u\sin\theta}{g} = \frac{2 \times 10 \times \sin{60{^\circ}}}{10} = \sqrt{3}s$
For vertical downward motion of particle $B$ from $B$ to $C$
We have
$h = \frac{1}{2}gT^{2} = \frac{1}{2} \times 10 \times \left( \sqrt{3} \right)^{2} = 15\ m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.