Basic kinematics of a balloon rising

In summary, the conversation discusses how to approach a physics problem involving a balloon rising at 10m/s and a passenger throwing a ball straight up at 12m/s. The conversation includes equations for calculating the time it takes for the passenger to catch the ball, as well as discussions on whether to account for gravity and how to consider the problem with respect to the ground. The final proposed solution is 2.44 seconds, but there is uncertainty as to whether the values given are already with respect to the ground.
  • #1
frasifrasi
276
0
Q- A balloon is rising at 10m/s when its passenger throws a ball straight up at 12m/s .How much later does the passenger catch the ball?

How would I approach this problem ? I am terrible at physics when it comes to doing the first basic step on a problem. If anyone can point me in the right direction it would be great.

Thank you!

Edit: units.
 
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  • #2
Think about doing the problem with respect to the speed of the balloon. (i.e. the ball is thrown at [tex]2 \tfrac{m}{s}[/tex] with respect to the balloon.)

And, in future, include units. It's really very inconvenient to have to figure out whether you meant the balloon started rising at t=10s or whether its speed at t=0 is [tex]10 \tfrac{m}{s}[/tex].
 
  • #3
what equaton should I use--should I take gravity into consideration?
 
  • #4
Ok, I have delta x = 0 = v_initial*t - 1/2*9.8*t^2.

which gives me t = 0 and t= .5s .
I used 2m/s for the velocity.


Is this correct.
 
  • #5
You should have gotten [tex] t = .41s.[/tex] Maybe you rounded wrong?
 
  • #6
Also, you could have solved the problem as a system of equations:
[tex]
\begin{align*}
y &= 10t\\
y &= 12t - \frac{1}{2}gt^2
\end{align*}
[/tex]​

Both ways result in the same answer.
 
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  • #7
the online homework site is saying that .41s (two sig figs) is not the right answer. Are u sure it is correct?
 
  • #8
frasifrasi said:
the online homework site is saying that .41s (two sig figs) is not the right answer. Are u sure it is correct?

I can't see how it wouldn't be.
 
  • #9
Do you use webassign?
 
  • #10
no, i use mastering physics. I copied and pasted the question, but it is giving me try again. what should I do?

fox, why didn't u take gravity into account for y=10t?
 
  • #11
well, I was assuming that it meant that the balloon was rising at constant speed. But, hey, why don't you try it with gravity? Maybe that'll be right. At least I hope so.
 
  • #12
but then the gravity will cancel out with the other gravity term.
 
  • #13
Hmm. you're right. Rrrr. Why don't you email your teacher. See if maybe mastering physics is wrong.
 
  • #14
What do you think of this:

speed of the balloon along vertical direction v = 10 m / s
speed of the ball along vertical direction u = 12 m / s
relative speed of the ball with respect to nallon V = 12 - 10 m / s
= 2 m / s
time taken to catch the ball by the passinger t = 2V /g
= 0.804 s
 
  • #15
2V/g = 2*2/9.81 = 0.41 s
 
  • #16
kamerling said:
2V/g = 2*2/9.81 = 0.41 s

That's what I got, but according to the website thingy he's using, that's not correct.
 
  • #17
frasifrasi said:
but then the gravity will cancel out with the other gravity term.
gravity should only 'apply' to the ball, since the gravity on the balloon is canceled by its upwards force..
the balloon travels at a constant 10 m/s, the ball starts at 12 m/s with acceleration -9.8 m/s².
 
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  • #18
OK, so I will talk to the ta and see what transpired...
 
  • #19
What about if I look at it relative to the ground:

The answer would be
d=10t for passenger,
d=22t-4.9t^2 (22m/s relative to the ground)

2.44s. Perhaps this is the answer--we will see. As much as I hate the long chapters and impossible problems on Halliday, at least it is easy to get a hold of the answers for that book.
 
  • #20
frasifrasi said:
What about if I look at it relative to the ground:

The answer would be
d=10t for passenger,
d=22t-4.9t^2 (22m/s relative to the ground)

2.44s. Perhaps this is the answer--we will see. As much as I hate the long chapters and impossible problems on Halliday, at least it is easy to get a hold of the answers for that book.

Hmm. I was thinking the values you were given were already with respect to the ground, but maybe only the balloon's speed was. In that case 2.44s would be the answer. But I'd talk to the ta anyway. The question should be clearer if that's so.
 

FAQ: Basic kinematics of a balloon rising

What is the principle behind a balloon rising?

The principle behind a balloon rising is the difference in density between the air inside the balloon and the surrounding air. The air inside the balloon is heated by a burner, making it less dense than the cooler air outside. This causes the balloon to rise and float in the air.

How does the shape of a balloon affect its rising?

The shape of a balloon affects its rising by impacting its aerodynamics. A spherical balloon has a lower drag coefficient and is more stable, making it easier to control and rise in the air. A cylindrical or irregularly shaped balloon may experience more turbulence and have a higher drag, making it harder to rise.

What factors influence the speed at which a balloon rises?

The speed at which a balloon rises is influenced by several factors, including the temperature difference between the inside and outside air, the size and shape of the balloon, and the weight of the balloon and its contents. The direction and strength of the wind can also affect the speed of a balloon's ascent.

How does altitude affect a balloon's performance?

As a balloon rises in altitude, the air becomes thinner and less dense. This can affect the balloon's buoyancy, as there is less air to displace and support its weight. It can also lead to a decrease in temperature, which can affect the performance of the burner and the balloon's lift capacity.

What are the safety considerations for a balloon rising?

Safety considerations for a balloon rising include proper training and experience for the pilot and crew, regular maintenance and inspections of the balloon and equipment, and monitoring weather conditions. It is also important to have emergency procedures in place and to follow all regulations and guidelines set forth by the Federal Aviation Administration (FAA).

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