Juggler Problem: Solving Kinematics for Ceiling Height

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To determine the minimum ceiling height for a juggler transferring balls between hands, the time each ball spends in the air must be calculated. The juggler's transfer time of 0.20 seconds can be interpreted as the time between tosses, assuming negligible overhead for throw and catch times. This means that each ball's flight time can be calculated using kinematic equations, focusing on the vertical component of motion. For juggling five balls released from a height of 1.5 meters, the ceiling height must accommodate the maximum height of the balls in flight. If the juggler increases to six balls, the ceiling height must be adjusted accordingly to ensure all balls can be successfully juggled.
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A juggler can transfer a ball from his left hand to his right hand in 0.20 s. If he is juggling 5 balls and releases them from a height of 1.5 m, what is the minimum height of the ceiling? How much higher should the ceiling be if he wants to juggle 6 balls ?

How are we going to solve this? If we use kinematics equations, we could problem get through it, but I am not sure how to.
 
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The first step is to figure out how long each ball spends in the air. (Not the 0.20 s part, but the part where the ball is tossed upward.)
 
Doesn't seem like there's enough information regarding catch and throw times for the juggled balls. If the transfer from left to right hand is a toss (what most jugglers do), then it's possble for the juggler to do the left hand transfer toss before the right hand does the upwards toss, using the .2 second window of transfer toss time to switch between the upward toss and catching the ball from the transfer toss. Same goes for the left hand catching the next ball during the .2 second window transfer toss time. This would reduce the time between tosses. On the other hand, the time between tosses could be greater than the .2 second toss window time.

Perhaps the idea is to assume throw and catch time overhead is effectively zero, so that the time between all tosses is .2 seconds.
 
rcgldr said:
Perhaps the idea is to assume throw and catch time overhead is effectively zero, so that the time between all tosses is .2 seconds.
That is how I am interpreting the problem.
 
All the information required to solve the problem is present, provided one makes some common-sense assumptions, including that 0.20 s is the time from catching a ball with one hand to throwing the same ball with the other. Another assumption would be that he can throw one ball at the same moment he catches the one behind it.

Thus, 0.20 s is simply dwell time at the bottom, as well as the time between each of the balls in flight. The only kinematic equation required is the one for the vertical component.
 
This might help in the visualization of the problem: http://www.flasharcade.com/arcade-games/juggling-simulator-game.html :)
 
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