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 9.87m/s.
angle = arctan(-momentum A/-mommentumB)
angle = -87.5
y = Vocsin(-87.5)+4.9t^2
i solved for t
and plugged t into
x = Voccos(-87.5)t
iam still getting an incorrect answer where am i going wrong?
The initial Height h of the whole thing is 45 meters and the velocity is zero since they say max height.
Therefore Part A is a chunk dropped from 45 meters
We want to know first the initial velocity of part A down, call it AV where v will be negative for down
note at AU, the horizontal speed of A is given 0 so it lands right under the explosion.
0 = 45 + AV t - 4.9 t^2
but t = .29 so
0 = 45 + AV(.29) -4.9 (.29)^2
0 = 45 +.29 AV - .4121
AV = -154 m/s and of course AU = 0
the initial vertical momentum of A (call it PAY) is .5*-154 = -77 kg m/s
Note PAX, the x momentum of A is zero)
Part B has no vertical momentum
Part C must have upward momentum of magnitude 77 to balance Part A downward momentum
Therefore PCY =+77
mass of Part C is 9.3 - .5 - 1 = 7.8 kg
and Vertical speed of Part C = 77/7.8 = 9.87 m/s
now Part B fell just like A did but with a constant horizontal speed
so it was in the air as long as Part A, .29 s
it went ten meters in .29 s
so its horizontal speed is 10/.29 = 34.5 m/s
Part B horizontal momentum = 34.5 m/s * 1 kg = 34.5 kg m/s
That must balance horizontal momentum of Part C
34.5 = 7.8 * CU
so CU = horizontal speed of C = 4.42 m/s
Now where does C land?
Initial speed up of Part C = 9.87
Initial height of part C = 45 m
I will split the part C vertical problem into going up and going down
going up
When at peak, v = 0
0 = 9.87 - 9.8 t
t rising = 1.01 s
h max = 45 + 9.87(1.01) - 4.9 (1.01)^2
h max = 50 m
falls from 50 m
50 = 4.9 (t falling)^2
t falling = 3.19
t = t rising + t falling = 4.2 seconds in the air
horizontal at 4.42 m/s for 4.2 s
so lands 18.6 m from Part A
now Part B fell WITH ZERO INITIAL SPEED DOWN and with a constant horizontal speed
How long was it in the air falling from 45 m?
45 = 4.9 t^2
t = 3.03 seconds
it went ten meters in 3.03 s
so its horizontal speed is 10/3.03 = 3.3 m/s
Part B horizontal momentum = 3.3 m/s * 1 kg = 3.3 kg m/s
That must balance horizontal momentum of Part C
3.3 = 7.8 * CU
so CU = horizontal speed of C = .423 m/s
Now where does C land?
Initial speed up of Part C = 9.87
Initial height of part C = 45 m
I will split the part C vertical problem into going up and going down
going up
When at peak, v = 0
0 = 9.87 - 9.8 t
t rising = 1.01 s
h max = 45 + 9.87(1.01) - 4.9 (1.01)^2
h max = 50 m
falls from 50 m
50 = 4.9 (t falling)^2
t falling = 3.19
t = t rising + t falling = 4.2 seconds in the air
horizontal at .423 m/s for 4.2 s
so lands 1.78 m from Part A
To find the distance at which Part C lands from Part A, you're on the right track by using the projectile motion equations. However, it seems like there might be a slight error in your calculations. Let's go through the steps together to identify where the mistake might be happening.
1. Calculate the momentum of Part A: Since Part A (0.5 kg) moves straight down, its momentum is given by p = m * v, where m is the mass and v is the velocity. The velocity of Part A can be found using the equation s = ut + 0.5at^2, where s is the distance (0.29 m), u is the initial velocity, a is the acceleration (taking as 9.8 m/s^2), and t is the time (0.29 s). Solving this equation will give you the initial velocity of Part A.
2. Calculate the momentum of Part B: Since Part B (1 kg) moves horizontally to the right, its momentum is given by p = m * v, where m is the mass and v is the velocity. The velocity of Part B is not given directly, but you can use the distance it travels horizontally (10 m) and the time it took to land (0.29 s) to calculate its horizontal velocity using the equation v = d/t.
3. Calculate the angle at which Part C moves: The angle at which Part C moves can be found using the equation angle = arctan(-momentum A / momentum B). Here, you need to make sure to divide the momentum of Part A by the momentum of Part B, keeping in mind their respective directions. Double-check if you used the correct signs for the momenta.
4. Calculate the time taken by Part C: To find the time taken by Part C, you can use one of the projectile motion equations in the vertical direction. Since the initial velocity of Part C is given (9.87 m/s), and the angle at which it moves is known, you can use the equation: y = Vo * sin(angle) * t - 0.5 * g * t^2. Solve this equation for t.
5. Calculate the distance of Part C from Part A: Finally, to find the distance at which Part C lands from Part A, you can use the horizontal projectile motion equation x = Vo * cos(angle) * t, where Vo is the initial velocity of Part C and t is the time calculated in step 4.
Double-checking each step and ensuring the correct use of signs and directions should help you arrive at the correct answer.