How many balls can this juggler juggle?

[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.

 

[Rake-MC]

Is this the only information you are given?

 

[zealot1985]

Yes that's all that's given really

 

[Rake-MC]

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

 

[zealot1985]

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.

 

[Rake-MC]

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..

 

[zealot1985]

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

 

[Borek]

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.

 

[zealot1985]

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?

 

[Rake-MC]

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:

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

and solve the quadratic.

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

 

[Rake-MC]

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

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.

 

[zealot1985]

So she can juggle 10 balls.

 

[Rake-MC]

Spot on!

 

[zealot1985]

Oh wicked. Stupid question this was.

 

[zealot1985]

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

 

[alphysicist]

So she can juggle 10 balls.

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.