Exploding particle?

Question

A projectile has mass 21 kg and is fired at a 60 degree angle above the horizontal with a speed of 80 m/s. At the highest point of its trajectory, it explodes into 2 pieces, one of which falls straight down with no initial speed. The 2 pieces have equal mass, and air resistance can be ignored. Find the distance from the point of firing that the other piece strikes assuming a level terrain, and find how much energy was released during the explosion.

Answer 1

I won't solve this but...
Find the vert & hor velocities
Vh = 80cos(60): Vv = 80sin(60)
Find the velocity of the other piece V1 from conservation of momentum, noting that the shell has only Vh at the apogee.
Find fall time = rise time t = Vv/g
Find the distance = Vh*t + V2*t
The kinetic energy added = KEf - KEi = (m/2)V1^2/2 - mVh^2/2
EDIT: The next answer errs in including ground impact speeds in kinetic energy. The energy added by the explosion only involves the velocities immediately before and after the explosion.

Answer 2

I had the full answer there in beginning, but shouldn't have done total work like that. You get the vertical velocity from 80 sin 60, and use that to get time to peak. t = V/9.8. Then use conservation of momentum to calculate second piece's new velocity. Note that at time of explosion, velocity is totally horizontal, and first piece loses all its horizontal and gains no vertical velocity.

For distance, use initial horizontal speed (80*cos 60) times time to top, plus horizontal speed after explosion times time to fall.

To get energy from explosion, is .5 mv^2 of first piece at contact with ground (will have vertical speed only) plus .5 mv^2 of second piece at contact with ground (will have vertical and horizontal speed, so use pythagorean theorem to get total speed). Then subtract initial energy .5(21)(80).