Post a New Question


posted by .

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 -

    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.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question