How Fast Must a Juggler Throw Four Balls?

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A juggler handling four balls takes 0.5 seconds to cycle each ball through his hands, which raises questions about the velocity needed to throw each ball and their positions during the juggling process. To maintain the juggling pattern, the juggler must throw each ball high enough to remain airborne for 1.5 seconds before catching it again. The discussion explores the timing of throwing and catching each ball, emphasizing the need for precise calculations of height and velocity. The juggler's technique may require adjustments if juggling five balls, as the time each ball spends in the air increases. Overall, the juggling dynamics hinge on understanding the physics of projectile motion.
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Homework Statement



A juggler is able to handle four balls simultaneously. He takes 0.5 s to cycle each
ball through his hands, throw the ball, and be ready to catch the next ball. (a)
With what velocity must he throw each ball up? (b) What is the position of the
other three balls when he has just caught one of the balls? (c) How high must he
throw the balls if he is to juggle five balls?

Homework Equations



Rotational kinematics eqns.
Rotational kinetic energy
etc.

The Attempt at a Solution



I'm not really sure.. A point in the right direction would help..
 
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EndoBendo said:

Homework Statement



A juggler is able to handle four balls simultaneously. He takes 0.5 s to cycle each
ball through his hands, throw the ball, and be ready to catch the next ball. (a)
With what velocity must he throw each ball up? (b) What is the position of the
other three balls when he has just caught one of the balls? (c) How high must he
throw the balls if he is to juggle five balls?

Homework Equations



Rotational kinematics eqns.
Rotational kinetic energy
etc.

The Attempt at a Solution



I'm not really sure.. A point in the right direction would help..

If he has just caught ball 1, he will throw that ball up in 0.5 seconds time.
At that same time [probably a fraction later] he will catch Ball 2

t= 0 catch ball 1
t = 0.5 throw ball 1 & catch Ball 2
t = 1.0 throw ball 2 & catch ball 3
t = 1.5 throw ball 3 & catch ball 4
t = 2.0 throw ball 4 & catch ball 1 again
etc.

So 1.5 seconds after throwing ball 1, he catches it again.

How fast does he have to throw it if it doesn't come back for 1.5 seconds?
How high will it get in that time?

Perhaps the idea of throwing one ball and catching the next at the same time if too far fetched.
Perhaps it is
t = 0 catch ball1
t = 0.5 throw ball 1
t = 1.0 catch ball 2
t = 1.5 throw ball 2 etc
.
.
.
t = 4.0 catch ball 1 again, so the ball has to be in the air for 3.5 seconds? For that to happen it would have to be one hell of a throw!
 
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