Question

A fire cracker is thrown with velocity of 30$m{s}^{-1}$ in a direction which makes an angle of ${75}^o$ with the vertical axis. At some point on its trajectory, the fire cracker split into two identical pieces in such a way that one piece falls 27 m far from the shooting point. Assuming that all trajectories are contained in the same plane, how far will the other piece fall from the shooting point ? (Take $g=10m{s}^{-2}$ and neglect air resistance)

• A. 63 m or 144 m
• B. 28 m or 72 m
• C. 72 m or 99 m
• D. 63 m or 117 m

Answer 1

Answer: (D)

63 m or 117 m

Given : $u=30$ m/s $m_1=m_2=m$

From figure, $\theta ={90}^o-{75}^o={15}^o$

Range of the centre of mass $R=\frac{u^{2}sin\left(2\theta \right)}{g}=\frac{(30{)}^2\times sin({30}^o)}{10}=45$ m

Using $x_{cm}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$ $⟹x_{cm}=\frac{x_1+x_2}{2}$

Center of mass of the system remains at point CM i.e $x_{cm}=45$ m

Case 1: If one of the pieces reaches point A i.e $x_1=27$

$\therefore$ $45=\frac{27+x_2}{2}$ $⟹x_2=63$ m

Case 2 : If one of the pieces reaches point A' i.e $x_1=-27$

$\therefore$ $45=\frac{-27+x_2}{2}$ $⟹x_2=117$ m