A rocket is projected straight up and explodes into three equally massive fragments just as it reaches the top of its flight (refer figure). One of the fragments is observed to come straight down in 2 sec, while the other two take 4 sec to come to ground, after the burst. Find the height h (in m) at which the fragmentation occurred.

Engineering

Physics

Collision

Question

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At the topmost point of the trajectory, the momentum of the system is zero. From conservation of momentum,

$m_{1} {\vec{v}}_{1} + m_{2} {\vec{v}}_{2} + m_{3} {\vec{v}}_{3} = 0$

as m₁ = m₂ = m₃

${\vec{v}}_{1} + {\vec{v}}_{2} + {\vec{V}}_{3} = 0$ .....(1)

The second and third fragments reach the ground simultaneously, therefore vertical components of v₂ and v₃ must be same; secondly, v₁ is downwards, the vertical components of v₂ and v₃ are $\frac{- v_{1}}{2}$ (i.e. directed upwards)
for first fragment, $h = v_{1} t_{1} + \frac{{gtt}_{1}^{2}}{2}$ ....(2)
for second fragment, $h = \frac{- v_{1} t_{2}}{2} + \frac{{gt}_{2}^{2}}{2}$ ....(3)
from equations (2) & (3), $v_{1} = \frac{g (t_{2}^{2} - t_{1}^{2})}{2 t_{1} + t_{2}}$
and $h = \frac{{gt}_{1} t_{2}}{2} (\frac{t_{1} - 2 t_{2}}{2 t_{1} + t_{2}})$

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