Explanation

(c) If a particle is projected with velocity $u$ at an angle $\theta$ with the horizontal, the velocity of the particle at the highest point is
$v = u\cos\theta = 200\cos{60{^\circ}} = 100\ ms^{- 1}$
If $m$ is the mass of the particle, then its initial momentum at highest point in the horizontal direction $= mv = m \times 100$. It means at the highest point, initially the particle has no momentum vertically upwards or downwards. Therefore, after explosion, the final momentum of the particles going upwards and downwards must be zero. Hence, the final momentum after explosion is the momentum of the third particle, in the horizontal direction. If the third particle moves with velocity $v^{'}$, then its momentum $= \frac{mv'}{3}$,
According to law of conservation of linear momentum,
We have $\frac{mv'}{3} = m \times 100$ or $v^{'} = 300\ ms^{- 1}$