Question

An object of mass $2m$ is projected with a speed of $100m{s}^{-1}$ at an angle $\theta ={\sin }^{-1}\left(\frac{3}{5}\right)$ to the horizontal. At the highest point, the object breaks into two 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, $g=10m/s^2)$

• A. 3840
• B. 1280
• C. 1440
• D. 960

Answer 1

Answer: (C)

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