Question

A firecracker 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 firecracker splits 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 place, how far will the other piece fall from the shooting point ? (Take g = 10 $m.s^{-2}$ 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

$\theta =15$

$Range=\frac{v^{2}sin2\theta }{g}=45m$

Since there is no horizontal force acting so the final CM should remain at 45m.

Let say the particle landed 27m away from initial point toward CM

Then $\frac{m\times 27+m\times x}{2m}=45\to x=63m$

if the object landed on other side of origin

$\frac{m\times -27+m\times x}{2m}=45\to x=117m$