Question

A particle $A$ is projected from the ground with an initial velocity of $10m{s}^{-1}$ at an angle of $60^{\circ }$ with horizontal. From what height should another particle $B$ be projected horizontally with velocity $5m{s}^{-1}$ so that both the particles collide in ground at point $C$ if both are projected simultaneously. (Take $g=10m{s}^{-2})$

• A. 10 m
• B. 15 m
• C. 20 m
• D. 30 m

Answer 1

Answer: (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
$h=\frac{1}{2}g{T}^2$
$=\frac{1}{2}\times 10\times (\sqrt{3}{)}^2=15m$