Basic kinematics

1. Feb 3, 2008

frasifrasi

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.

Last edited: Feb 3, 2008
2. Feb 3, 2008

foxjwill

Think about doing the problem with respect to the speed of the balloon. (i.e. the ball is thrown at $$2 \tfrac{m}{s}$$ 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 $$10 \tfrac{m}{s}$$.

3. Feb 3, 2008

frasifrasi

what equaton should I use--should I take gravity into consideration?

4. Feb 3, 2008

frasifrasi

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. Feb 3, 2008

foxjwill

You should have gotten $$t = .41s.$$ Maybe you rounded wrong?

6. Feb 3, 2008

foxjwill

Also, you could have solved the problem as a system of equations:
\begin{align*} y &= 10t\\ y &= 12t - \frac{1}{2}gt^2 \end{align*}​

Both ways result in the same answer.

Last edited: Feb 3, 2008
7. Feb 3, 2008

frasifrasi

the online homework site is saying that .41s (two sig figs) is not the right answer. Are u sure it is correct?

8. Feb 3, 2008

foxjwill

I can't see how it wouldn't be.

9. Feb 3, 2008

foxjwill

Do you use webassign?

10. Feb 3, 2008

frasifrasi

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. Feb 3, 2008

foxjwill

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. Feb 3, 2008

frasifrasi

but then the gravity will cancel out with the other gravity term.

13. Feb 3, 2008

foxjwill

Hmm. you're right. Rrrr. Why don't you email your teacher. See if maybe mastering physics is wrong.

14. Feb 4, 2008

frasifrasi

What do you think of this:

speed of the ballon 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. Feb 4, 2008

kamerling

2V/g = 2*2/9.81 = 0.41 s

16. Feb 4, 2008

foxjwill

That's what I got, but according to the website thingy he's using, that's not correct.

17. Feb 4, 2008

timon

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.

Last edited: Feb 4, 2008
18. Feb 4, 2008

frasifrasi

OK, so I will talk to the ta and see what transpired...

19. Feb 4, 2008

frasifrasi

What about if I look at it relative to the ground:

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. Feb 4, 2008

foxjwill

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.