Problem 14.15 A 13-kg projectile is passing through the origin O with a velocity v0 = (35 m/s)i when it explodes into two fragments A and B, of mass 5 kg and 8 kg, respectively. Knowing that 3 s later the position of fragment A is (90 m, 7 m, –14 m), determine the position of fragment B at the same instant. Assume 9.81m/s y a g   2 and neglect air resistance

(114.38, -4.37, -63)m

Explanation:

Let be the velocities of object A and B at the instant right after the explosion, respectively. If we ignore air resistance and take gravity in the z direction into account then the equation of motion for object A can be written as

We can apply the law of momentum conservation during the explosion to solve for the initial velocity of B

From here we can derive the 3 equation at 3 directions

1. In the x direction:

So after 3s object B would be at 38.13*3 = 114.38 m in the x-direction

2. In the y direction:

So after 3s object B would be at -1.456*3 = -4.37 m in the y direction

3. In the z direction:

So after 3 s object B would be at

it was a fall down when it happens

explanation: