A projectile of mass 2m is projected at an angle of 45 degrees with the horizontal with a velocity 20root[2]m/s. After 1 s explosion takes place and the particle is broken into two equal pieces. As a result of explosion one part comes to rest. Find the maximum height attained by the other part. Take g=10m/s^2.

The answer for this question is 35 m.

Initially the velocity is 20root(2)ms^-1 & the angle is 45. Vertical component of velocity = 20ms^-1

After 1 s, It will cover a distance of s = ut+1/2at² = (20)(1)+1/2(-10)(1)² = 20 - 5 = 15m and its vertical velocity comp. will be reduced to 10ms^-1.

Soon After it s mass gets reduced to m from 2m due to reaking so according to the law of conser. of center of mass,  it velocitys vertical comp. will be doubled. = 10ms^-1.*2 = 20ms^-1

Now for coming to rest it has to cover diatnce v²= u²-2gs=> 400/20=s = 20m

So the total height achieved by the the ball is 20+15 = 35 m