$A$ projectile moving vertically upwards with a velocity of $200\, ms^{-1}$ breaks into two equal parts at a height of $490\, m$. One part starts moving vertically upwards with a velocity of $400\, ms^{-1}$. How much time will it take,after the break up,for the other part to hit the ground? .............. $s$

$A$ spacecraft of mass $M$ is moving with velocity $V$ and suddenly explodes into two pieces. $A$ part of it of mass $m$ comes to rest. The velocity of the other part will be:

$A$ projectile of mass $m$ is projected from ground with a speed of $50 \, m/s$ at an angle of $53^{\circ}$ with the horizontal. It breaks up into two equal parts at the highest point of the trajectory. One particle comes to rest immediately after the explosion. The distance between the pieces of the projectile when they reach the ground is: (in $, m$)

The figure shows a man of mass $m$ standing at the end $A$ of a trolley of mass $M$ placed at rest on a smooth horizontal surface. The man starts moving towards the end $B$ with a velocity $u_{rel}$ with respect to the trolley. The length of the trolley is $L$. The time taken by the man to reach the other end is

$A$ body of mass $5 \ kg$ initially at rest explodes into three fragments. The ratio of the masses of the three fragments is $1 : 1 : 3$. The two fragments of equal mass move in mutually perpendicular directions with a velocity of $21 \ m/s$. What is the velocity of the heaviest fragment in $m/s$?

$A$ body is moving with a velocity $v$ and breaks up into two equal parts. One of the parts retraces its path with velocity $v$. Then the velocity of the other part is: