A shell of mass 2 m fired with speed u at an angle θ to the horizontal explodes at the highest point of its motion into two fragments of mass m each. If one fragment, whose initial speed is zero, falls vertically, .the distance at which the other fragment falls from the gun is given by

Momentum of the shell at highest point=2m(ucos⁡θ)As one fragment has initial speed zero (stationary) and hence the other fragment of mass m has velocity component (2ucos⁡θ) in horizontal direction. This will move under gravity. So it wilt fall at a distance which is equal to range.R=u2sin⁡2θgHence, distance covered from the gun=R2+R=32R=32u2sin⁡2θg