Question

An object of mass $2\, m$ is projected with a speed of $100\, ms ^{-1}$ 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 $\left(\right.$ Given, $\left. g=10\, m / s ^{2}\right)$

• 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
$2 m u \cos \theta=m(0)+m v$
or $v=2 u \cos \theta$
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
$=\frac{R}{2}+R=\frac{3 R}{2}$
$=\frac{3}{2} \frac{u^{2} \sin 2 \theta}{g}=\frac{3}{2} \times \frac{(100)^{2} \times 2 \sin \theta \cos \theta}{g}$
$-\frac{3}{2} \times \frac{(100)^{2} \times 2 \times \frac{3}{5} \times \frac{4}{5}}{10}-1440\, m$