# 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: