Posted by **seth** on Monday, December 22, 2008 at 1:33pm.

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 got part C's initial velocity as 10.05m/s.

Now i want to use y = 10.05sinAt + 4.9t^2

to find t

then use t to find the distance. But how do i find the angle A part C moves at?

- physics -
**drwls**, Monday, December 22, 2008 at 4:56pm
The angle that part C moves at can be figured out by applying a momentum balance.

Initial skyrocket momentum at maximum height = 0

= sum of the momenta of parts A, B and C

The mass of part C is 9.3 - 0.5 - 1.0 = 7.8 kg. Its upward momentum component is the same as part A's but in the opposite direction (up). Its horizontal momentum component is the same as part B's, but to the left.

I am neglecting the momentum of gases released in the separation explosion, but this is probably OK.

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