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)?
The Attempt at a Solution
can you check my work
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.
momentum A = 78.2967
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 = 10/t = 10/3.03 = 3.3m/s
momentum B = (1kg) (Vx) = 3.3
3. Resultant of the two momentums:
Did 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.
momentum C =(momA ^2 + momB ^2) ^0.5 = 78.3662
Divide the momentum of C by its mass to get the initial velocity of C.
Voc = 10.046m/s
i am stuck at this point please help thanks