Posted by seth on .
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?
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.