Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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