Two balls of equal masses are thrown upwards along the same vertical direction at an interval of 2 s with same initial velocity of 39.2 ms^–1. Then find the height at which they collide.

Given, that the balls are identical i.e. have the same mass.

We call the ball thrown at t=0 s as ball 1 and the ball thrown at t= 2s as ball 2.

Both the balls have same mass, same initial speed, same acceleration (–g), so both will undergo the same kind of motion.

The only difference is that one is thrown at t= 0s and other is thrown at t= 2s. So ball 2 will cover the same distance 2 seconds later

Let time taken by ball 1 to reach height h = t

Tehn time taken by ball 2 to reach height h = t–2

Using,

For ball 1,

For ball 2

Equating both the values of h,

Putting value of t in

We get,