An object of mass $2\ m$ is projected with a speed of ${100\ ms}^{- 1}$ at angle $ heta = \sin^{- 1}\left( \frac{3}{5} ight)$ to the horizontal. At the highest point, the object breaks into pieces of same mass $m$ and the first one comes to rest. The distance between the point of projection and the point of landing of the bigger piece (in metre) is $(given,\ g = {10\ ms}^{- 2})$

Question

An object of mass $2\ m$ is projected with a speed of ${100\ ms}^{- 1}$ at angle $	heta = \sin^{- 1}\left( \frac{3}{5} ight)$ to the horizontal. At the highest point, the object breaks into pieces of same mass $m$ and the first one comes to rest. The distance between the point of projection and the point of landing of the bigger piece (in metre) is $(given,\ g = {10\ ms}^{- 2})$

Dr. Abubacker · Accepted Answer

(c) Horizontal range of the object fired,
$R = \frac{u^{2}\sin{2	heta}}{g}$
At the highest point, when object is exploded into two equal masses, then
$2mu\cos{	heta = m(0) + mv}$
$or\ \ v = 2u\cos	heta$
It means, the horizontal velocity becomes double at the highest point, hence it will cover double the distance during the remaining flight.
$	herefore Total\ horizontal\ range\ of\ the\ other\ part$
$= \frac{R}{2} + R = \frac{3R}{2}$
$= \frac{3}{2}\frac{u^{2}\sin{2	heta}}{g}$
$= \frac{3}{2} 	imes \frac{(100)^{2} 	imes 2\sin{	heta\cos	heta}}{g}$
$= \frac{3}{2} 	imes \frac{(100)^{2} 	imes 2 	imes \frac{3}{5} 	imes \frac{4}{5}}{10} = 1440\ m$

Explanation

Reviewed by Dr. Abubacker

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