Posted by **seth** on Sunday, December 21, 2008 at 6:42pm.

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)?

- physics -
**drwls**, Sunday, December 21, 2008 at 8:28pm
I will be happy to critique your work, when shown. This is an exercise in applying Newton's second law.

- physics -
**GK**, Sunday, December 21, 2008 at 9:18pm
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.

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