An object of mass $2 m$ is projected with a speed of $100 \mathrm{~ms}^{-1}$ at an angle $\theta=\sin…
An object of mass $2 m$ is projected with a speed of $100 \mathrm{~ms}^{-1}$ 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
(Given, $g=10 \mathrm{~m} / \mathrm{s}^2$ )
$3840$
$1280$
$1440$
$960$
Solution
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
$\begin{aligned} & =\frac{R}{2}+R=\frac{3 R}{2} \\ & =\frac{3}{2} \frac{u^2 \sin 2 \theta}{g}=\frac{3}{2} \times \frac{(100)^2 \times 2 \sin \theta \cos \theta}{g} \\ & =\frac{3}{2} \times \frac{(100)^2 \times 2 \times \frac{3}{5} \times \frac{4}{5}}{10}=1440 \mathrm{~m}\end{aligned}$