A shell is fired from a cannon with a velocity V at an angle `theta` with the horizontal direction. A the highest point i its path, it explodes into two pieces of equal masses. One of the pieces retraces its path to the cannon. The speed of the other priece immediately after the explocison is

Explanation

The velocity of the shell at the highest point of tarajector is `v_(M)=ucostheta=100 cos 60^(@)=50 m//s`
Let `v_(1)` be the speed of the fragment which moves long the negative `x`-direction and the other fragment has speed `v_(2)`, which must along the positive `x`-direction. now from, momentum conservation, we have
`mv=(-m)/2 v_(1)=m/2v_(2)` or `2v=v_(2)-v_(1)`
or `v_(2)=2v+v_(1)=(2xx50)+50=150m//s`