What Are the Physics Behind a Juggler's Free-Falling Balls?

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The discussion focuses on the physics of juggling, specifically the calculations involved when a juggler throws balls vertically in a room with a ceiling height of 4.2 meters. To determine the initial velocity required for the ball to reach the ceiling, the equation v² = u² + 2as is utilized, where 'v' is the final velocity (0 m/s at the ceiling), 'u' is the initial velocity, 'a' is the acceleration due to gravity (-9.8 m/s²), and 's' is the distance (4.2 m). The time taken for the ball to reach the ceiling can be calculated using the same equation, and the second ball's trajectory is analyzed to find when both balls pass each other.

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A juggler performs in a room whose ceiling is 4.2 m above the leve of this hands. He throws a ball vertically upward so that it just reaches the ceiling.
a. with what initial velocity does he throw the ball?
b. what time is required for the ball to reach the ceiling.

He throws a second ball upward with the same initial velocity at the instant the first ball is at the ceiling.
c. how long after the second ball is thrown do the two balls pass each other?
d. when the balls pass each other, how far are they above the juggler's hand?
i understand how to do the equations and everything, i just don't know which apply with which? if someone could help me out be telling me which equations i would use, that would be great :)
 
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best way to do this is to write out what you have:

s=4.2
u=?
v=0
a=-9.8
t=?

for Q1, you need to find u

the three equations you have are:

a) v=u+at
b) s=ut +1/2 a t^2
c) v^2=u^2 + 2as

you can't use a) or b) as you don't yet know t

so i'd say c) would work

do the same for your other questions i guess...
 

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