help please with physics?

Question

A 1457.0 kg weather rocket accelerates upward at 5.20 m/s2. It explodes 4.38 s after liftoff and breaks into two fragments, one twice as massive as the other. Photos reveal that the lighter fragment travelled straight up and reached a maximum height of 559.0 m measured from the ground. What was the velocity and direction of the heavier fragment just after the explosion? (Assume the positive direction is upward.)

Answer 1

s = s0 + v0t + (1/2)at^2
v = v0 + at
At the time of the explosion
v = 4.38*5.2 = 22.776 m/s
s = 0.5*5.2*4.38^2 = 49.87944 m
After the explosion, the velocity of the smaller fragment was
v = √(2*9.80665(559 - 49.87944)) = 99.928 m/s
To find the velocity of the larger fragment,
1,457*22.776 = 1,457*99.928/3 + 2*1,457v/3
22.776 = 99.928/3 + 2v/3
2v = 68.328 - 99.928 = -31.600 m/s
v = - 15.800 m/s or 15.8 m/s straight down.