A particle $A$ is projected from the ground with an initial velocity of $10\ ms^{- 1}$at an angle of 60${^\circ}$ with horizontal. From what height $h$ should an another particle $B$ be projected horizontally with velocity $5ms^{- 1}$ so that both the particles collide in ground at point $C$ if both are projected simultaneously? $(g = 10\ ms^{- 2})$ [image:4_image44.emf]

Question

Dr. Abubacker · Accepted Answer

(c) Horizontal component of velocity of $A$ is 10 $cos\ 60{^\circ}$ or $5\ ms^{- 1}$ which is equal to the velocity of $B$ in horizontal direction. They will collide at $C$ if time of flight of the particles are equal or $t_{A} = t_{B}$
$\frac{2u\sin	heta}{g} = \sqrt{\frac{2h}{g}}\ \ \ \ \ \ \ \ \ \left( 	herefore h = \frac{1}{2}gt_{B}^{2} ight)$
or $h = \frac{2u^{2}\sin^{2}	heta}{g}$
$= \frac{2(10)^{2}\left( \frac{\sqrt{3}}{2} ight)^{2}}{10} = 15\ m$

Explanation

Reviewed by Dr. Abubacker

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