Answer:
We can start by using conservation of momentum and conservation of energy to solve this problem.
Conservation of momentum:
Before the explosion, the momentum of the projectile is:
p = mv = (228 kg)(146 m/s) = 33312 kg·m/s
After the explosion, the momentum is still conserved, so the momentum of each piece must be equal to 1/3 of the initial momentum:
p = (1/3) mv1 + (1/3) mv2 + (1/3) mv3
where v1, v2, and v3 are the velocities of the three pieces after the explosion.
Conservation of energy:
At the highest point of the arc, the projectile has no kinetic energy and all of its energy is in the form of potential energy. Therefore, we can use conservation of energy to find the potential energy at this point and use that to find the initial kinetic energy of the projectile.
mgh = (1/2) mv²
where h is the maximum height of the projectile and v is the initial velocity of the projectile.
Solving for v, we get:
v = sqrt(2gh)
where g is the acceleration due to gravity (9.81 m/s²).
Substituting the given values, we get:
v = sqrt(2(9.81 m/s²)(h)) = 146 m/s
Now we can use the conservation of momentum equation to solve for the velocity of the third fragment:
p = (1/3) mv1 + (1/3) mv2 + (1/3) mv3
33312 kg·m/s = (1/3)(228 kg)(146 m/s) + (1/3)(228 kg)(146 m/s) + (1/3)(228 kg)(v3)
Simplifying and solving for v3, we get:
v3 = 306.23 m/s
Therefore, the magnitude of the velocity of the third fragment immediately after the explosion is 306.23 m/s.
To determine the direction of the velocity of the third fragment immediately after the explosion, we can use the fact that two of the fragments move with the same speed right after the explosion as the entire projectile had just before the explosion. This means that the horizontal component of the velocity of the third fragment must be equal in magnitude to the horizontal component of the velocity of the original projectile (146 m/s), and the vertical component must be opposite in sign to the vertical component of the velocity of the fragment that moves vertically downward.
Since the fragment that moves vertically downward has a velocity that is purely vertical, its vertical component is equal in magnitude to its total velocity. Therefore, the vertical component of the velocity of the third fragment must also be equal in magnitude to 306.23 m/s.
Using the Pythagorean theorem, we can find the magnitude of the horizontal component of the velocity of the third fragment:
sqrt((146 m/s)² - (306.23 m/s)²) = 259.31 m/s
Therefore, the velocity of the third fragment immediately after the explosion has a magnitude of 306.23 m/s and is directed at an angle of arctan(-306.23 m/s / 259.31 m/s) = -51.4° below the horizontal.