How Fast Must a Juggler Throw Four Balls?

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SUMMARY

The juggler must throw each ball with a velocity sufficient to keep it airborne for 1.5 seconds to maintain a four-ball juggling pattern. Using the equation for free fall, the required initial velocity can be calculated as approximately 14.7 m/s. Additionally, when the juggler catches one ball, the other three balls will be at various heights, determined by their respective times in the air. To successfully juggle five balls, the juggler must increase the throw height, necessitating a higher initial velocity.

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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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