Solving the Air Balloon & Pellet Problem

In summary, a hot air balloon and a pellet fired from the ground both have the same altitude at two different points. Using the equation vt=Vt+0.5(-9.81)t^2, and finding the two times where the distance is the same, it can be determined that these points are 16.207 meters and 16.193 meters above the ground level.
  • #1
Kandy
27
0
A hot air balloon is ascending straight up at a constant speed of 7.0 m/s. When the balloon is 12.0m above the ground, a gun fires a pellet straight up from the ground level with an initial speed of 30.0 m/s. Along the paths of the balloon and the pellet, there are two places where each of them has the same altitude at the same time. How far above the ground level are these places?
For this problem, I need to find two distances, so I tried to find the two times for the distances first but I can't work it out. This is what I have:
let v=velocity of balloon, t=time, V=velocity initial, a=acceleration
vt=Vt+0.5(-9.81)t^2
7t=30t+(-4.91)t^2
7t=30t-4.91t^2
0=23t-4.91t^2
then I factored it
0=t(23-4.91t)
t=0, 4.68 or 4.7
I'm missing one more time and distance or is this time also incorrect? I don't know how to solve this problem, please help me out :redface:
 
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  • #2
you need to decide WHEN you're going to start timing ...
either when the balloon leaves the ground,
or when the bullet leaves the ground.

I would start the timer when the pellet is fired.
Where is the balloon when MY stopwatch reads zero?
 
  • #3
I redid the problem but I still can't get it because the two times are the same. What did I do wrong now?

vt-12.0=Vt+0.5(-9.81)t^2
7.0t-12.0=30.0t+(-4.91)t^2
7.0t-12.0=30.0t-4.91t^2
0=-4.91t^2+30.0t+12.0

P=-58.9
S=30.0

(-4.91t^2+31.85t)(-1.85t+12.0)
t(-4.91t+31.85)(-1.85t+12.0)

t=0

-4.91t+31.85=0
-4.91t=-31.85
t=6.49

-1.85t+12.0=0
-1.85t=-12.0
t=6.49
 
  • #4
1) why did you write vt - 12? When the bullet is fired the balloon is 12m over the ground.
2) The initial speed of the bullet is 30 + 7 (which is the speed of the balloon) - unless your using some other referance system.
3) Don't forget to add the bullets initial position to it's position equation.
 
  • #5
i don't understand why the speed of the balloon needs to be added to the initial speed of the bullet.
 
Last edited:
  • #6
Kandy,

you should have 12[m] + 7[m/s] t = 30[m/s] t - ... ,
which SHOULD result in 0 = -12[m] + (30 - 7)[m/s] t ...

your writing would be a lot easier to read
if you started sentences with the SUBJECT

y(balloon) = 12[m] + v_i t
y(pellet) =

Don't try to write one very complicated sentence
that includes all the important information in some hidden form!
Write a few small statements, instead, so you won't get confused.
(and you're allowed to use words to explain).
 
Last edited:
  • #7
12.0[m] + 7.0[m/s] t = 30.0[m/s] t - 0.5 (-9.81[m/s^2]) t^2
0 = -4.91[m/s^2] t^2 + 23.0[m/s] - 12.0[m]

factor

product=59 sum=23.0

the two numbers are 2.95 and 20.05

(-4.91 t^2 + 2.95t) (20.05 t - 12.0)
t (4.91 t + 2.95) (20.05 t - 12.0)

t=0

4.91 t + 2.95 = 0
4.91 t = -2.95
t = 0.601

20.05 t - 12.0 = 0
20.0 t = 12.0
t= 0.599

now I substitute the times in for distance

12.0[m] + 7.0[m/s] t
12.0[m] + (7.0[m/s]) (0.601)
12.0[m] + 4.207[m]
= 16.207[m]

12.0[m] + 7.0[m/s] t
12.0[m] + (7.0[m/s]) (0.599)
12.0[m] + 4.193[m]
= 16.193[m]

That's only a difference of 2cm (which makes me feel like it's incorrect) but is this correct. Plz let it be
 
  • #8
Kandy said:
12.0[m] + 7.0[m/s] t = 30.0[m/s] t - 0.5 (-9.81[m/s^2]) t^2
0 = -4.91[m/s^2] t^2 + 23.0[m/s] - 12.0[m]
...
This is correct.

You should now just use the quadratic formula.

you have,

at² + bt - c = 0

t = {-b +/- sqrt(b² - 4ac)}/2a

You should get t = 0.6s and t = 4s (approx)
 
  • #9
I got the same answer (0.6 and 4) now, but I used a graphing calculator; I don't know how to get to t={-b+/- sqrt... because I haven't learned how yet.
 

1. How do you solve the air balloon and pellet problem?

The air balloon and pellet problem can be solved by using the ideal gas law, which states that the pressure, volume, and temperature of a gas are related by the equation PV = nRT. By manipulating this equation and using the given information about the balloon and pellet, the final volume of the balloon can be determined.

2. What is the ideal gas law?

The ideal gas law is a mathematical equation that describes the behavior of an ideal gas. It states that the pressure, volume, and temperature of a gas are directly proportional to each other, with the constant R representing the gas constant.

3. What is the purpose of solving the air balloon and pellet problem?

The purpose of solving this problem is to determine the final volume of the air balloon after the pellet has been released into it. This can help in predicting the behavior of gases and understanding the relationship between pressure, volume, and temperature.

4. What factors affect the final volume of the air balloon?

The final volume of the air balloon is affected by the initial volume of the balloon, the amount of air in the balloon, the temperature of the air, and the volume of the pellet. These factors can be manipulated using the ideal gas law to determine the final volume.

5. What are some real-world applications of solving the air balloon and pellet problem?

Some real-world applications of solving this problem include predicting the behavior of gases in different environments, such as in hot air balloons or in chemical reactions. It can also be used in engineering to determine the optimal volume for gas-filled containers. Additionally, understanding the ideal gas law can help in understanding the effects of temperature and pressure on gas behavior in various industries, such as in the production of food and beverages.

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