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

To find the magnitude of the momentum of the 5 lb piece just before it strikes the ground, we'll first need to determine its horizontal velocity.

We know that the time it takes for an object to reach its maximum height when launched vertically is the same as the time it takes to fall back down to the ground. This means that the total time of flight (t) for both pieces is the same.

Using the kinematic equation for vertical motion, we can determine the total time of flight. The formula is:

H = (1/2) * g * t^2

Where H is the maximum height, g is the acceleration due to gravity (which is approximately 32 ft/s^2), and t is the time of flight.

Plugging in the given values:
837 ft = (1/2) * 32 ft/s^2 * t^2

Solving this equation will give us the value of t.

Now, let's focus on the horizontal motion. We are given that the 5 lb piece hits the ground 1,564 ft away from the explosion. We can use the formula for horizontal velocity to find its magnitude.

Vx = d / t

Where Vx is the horizontal velocity, d is the horizontal distance traveled, and t is the time of flight.

Plugging in the given values:
Vx = 1564 ft / t

Now that we have the horizontal velocity, we can find the momentum (p) of the 5 lb piece.

Momentum is defined as the product of an object's mass and its velocity:
p = m * Vx

Where m is the mass of the 5 lb piece and Vx is the horizontal velocity.

Plugging in the given mass and the calculated horizontal velocity, we can find the magnitude of the momentum.