# How many balls can this juggler juggle?

1. Oct 13, 2008

### zealot1985

A juggler has a set of balls that she can toss into the air. If it takes her 0.2 s to transfer a ball from the receiving hand to the launch hand and she is able to throw a given ball up with a maximum speed of 10 m/s how many balls can she successfully juggle at any time? Assume that the balls are traveling in a straight line but do not collide.

Last edited: Oct 13, 2008
2. Oct 13, 2008

### Rake-MC

Re: Question

Is this the only information you are given?

3. Oct 13, 2008

### zealot1985

Re: Question

Yes that's all that's given really

4. Oct 13, 2008

### Rake-MC

Re: Question

Is there a diagram included? This question confuses me because it says the balls travel in a straight line..

5. Oct 13, 2008

### zealot1985

Re: Question

No diagram. That's all that's given. My professor handed this question to us and I'm just as confused. No mention of a parabola-form or anything.

6. Oct 13, 2008

### Rake-MC

Re: Question

Well.. You will want to get someone else's opinion on this, but I would just say two balls. She will throw one at 10ms-1 to her catching hand, and at the same time throw one from the catching hand to the other hand. They both travel at the same speed and the exchange will take 0.2s as defined...

However I'm sure we've overlooked something huge..

7. Oct 13, 2008

### zealot1985

Re: Question

This question's been worded really badly. It's ridiculous.

I think all the balls are forming a parabola path. The launch speed is 10ms-1

8. Oct 13, 2008

### Staff: Mentor

Re: Question

Question is OK. You have to calculate how long ball thrown up with initial speed 10 m/s will be in the air, that'll let you calculate how many balls can be processed by the juggler before she has to deal with the first one again.

Last edited: Oct 13, 2008
9. Oct 13, 2008

### zealot1985

Re: Question

Trip up:
Vf = Vi + at
For the trip up, final velocity = 0, since it stops at the max height.

0= 10 - 10t
t = 1s

Trip down:
Same thing, so it is 2 seconds for the entire trip?

10. Oct 13, 2008

### Rake-MC

Re: Question

Oh ok I just understood it. So it IS thrown in a straight line, it's just completely vertical.
So use constant acceleration formulae to calculate the time it's in the air for.
You can either use:

$$v = u + at$$

and double that answer, or use:

$$s = ut + \frac{1}{2}at^2$$

Key: u = initial velocity
v = final velocity
t = time
a = acceleration
s = displacement

11. Oct 13, 2008

### Rake-MC

Re: Question

Yeah that sounds fine, in most courses, acceleration due to gravity is measured as 9.8ms-2 or 9.81ms-2. However, if you have been dealing with it as 10ms-2 for your course then it's best to leave it.

12. Oct 13, 2008

### zealot1985

Re: Question

So she can juggle 10 balls.

13. Oct 13, 2008

### Rake-MC

Re: Question

Spot on!

14. Oct 13, 2008

### zealot1985

Re: Question

Oh wicked. Stupid question this was.

15. Oct 13, 2008

### zealot1985

Re: Question

Thank you very much. I'm done for the night.
I'll have another question to work upon for tomorrow though.

16. Oct 13, 2008

### alphysicist

Re: Question

I agree with the kinematics calculations, but I don't believe this is the answer.

For example, if the time of flight had been 0.2 seconds (instead of 2 seconds), she would have been able to juggle two balls, not one.