Physics
posted by Charlie .
A cannon shoots a shell straight up. It reaches its maximum height, 837 feet, and splits into two pieces, one weighing 2 lb and the other 5 lb. The two pieces are observed to strike the ground simultaneously. The 5 lb piece hits the ground 1,564 feet away from the explosion (measured along the x axis). Find the magnitude of the momentum of the 5 kg piece just before it strikes the ground.

How long did it take to fall?
t=sqrt(2d/g)
What was the horizontal momentum:
momentum=mass(1564/time)
Now find the vertical momentum, and add them as vectors. 
Since they arrive at the same time, no vertical momentum is imparted by the explosion. Use the vertical distance that the 5 lb piece travels down from 837 ft, and the horizontal distance of 1564 ft, to get its horizontal velocity. Since momentum is conserved, the 2 lb piece acquires 2.5 times the horizontal velocity of the 5 lb piece but in the opposite direction. The vertical velocity component when it hits the ground is sqrt (2 g * 837 ft).
Use the vertical and horizontal components of V to get the megnitude. 
It is falling from 837 feet
Do the vertical problem first
v = 32 t
d = 837 = (1/2) a t^2 = 16 t^2
so t^2 = 52.3
t = 7.23 seconds
v at ground = 32 (7.23) = 231 ft/s
now call horizontal speed u
u is constant because there is no force in the horizontal direction
so speed = distance/time
u = 1564 ft/7.23 seconds = 216 ft/s
magnitude of speed = sqrt (u^2 + v^2) = sqrt (216^2 + 231^2) = sqrt(46795+53568)
= 316 ft/s
multiply that by the mass to get magnitude of momentum
Now I am not clear about the mass of this piece, 5 pounds or 5 kilograms
if it is five pounds use 5/32 for mass in slugs
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