Physics

$p =mv$

$F=dtdp $

Since $F=0$

Therefore $p=$ constant

When the particles collide the height of both particles will be same,

For first particle time will be $t$ and for second particle time will be $t−2$

$h=ut−21 gt_{2}=u(t−2)−21 g(t−2)_{2}$

$ut−21 gt_{2}=ut−2u−21 g(t_{2}−4t+4)$

$0=−2u+2gt−2g$

$t=gu +1$

$t=5s$

By solving the quadratic equation in $t$ we get the time of first collision to be $5$s.

$h=39.2×5−21 g×5_{2}=20g−12.5g=7.5gmeter$

Now by momentum conservation, velocity of particles after collision

$m(u−gt)+m(u−g(t−2))=mv_{′}$

$mu−5mg+mu−3mg=mv_{′}$

$v_{′}=2u−8g=0m/s$

$a=−g$

Now for collision with ground let time taken be $t_{2}$

$h=21 gt_{2}⇒t=g2h =g2×7.5g =15 $

Thus, times of total time of flights will be $5+15 $s and $3+15 $s