posted by seth .
A 9.3 kg firework is launched straight up and at its maximum height 45 m it explodes into three parts. Part A (0.5 kg) moves straight down and lands 0.29 seconds after the explosion. Part B (1 kg) moves horizontally to the right and lands 10 meters from Part A. Part C moves to the left at some angle. How far from Part A does Part C land (no direction needed)?
I will be happy to critique your work, when shown. This is an exercise in applying Newton's second law.
1. Momentum for part A:
y = (Vo)t - (1/2)(9.8m/s^2)t^2
Substitute the values for y and t which are given. Solve for Vo, the initial downward velocity. Multiply the value of Vo by 0.5 kg to get the initial momentum down of A.
2. Momentum for part B:
y = (1/2)gt^2 = 45m for its downward motion.
substitute the value of g and solve for t.
Vx = 10m/t
Now you can calculate the momentum of B
3. Resultant of the two momentums:
Do a vector addition of the momentums of A and B. The resultant will be a vector downward to the right. Determine both, magnitude and direction.
4. Momentum of part C:
That is opposite (equilibrant) of of the vector sum of A and B. Determine it.
Divide the momentum of C by its mass to get the initial velocity of C. That will make part C a projectile at an upward angle. It will be tricky solving this part since it lands 45 meters below the point of "launching".
Best of luck.