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
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?
2 answers
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
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
2 answers