Juggler throws 2 balls - when do they pass

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A juggler throws two balls, each at 10 m/s, with the first reaching a height of 2.4 m in 0.24 seconds. The second ball is thrown as the first reaches the ceiling. To determine when the two balls pass, it is established that the time after the second ball is thrown is 0.12 seconds. This is calculated using the equation 20x = 2.4 m, where x represents the time after the second ball is thrown. The solution effectively demonstrates the timing of the balls' paths in relation to each other.
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Homework Statement



A juggler throws a ball at 10m/s and it takes 0.24 seconds to reach the ceiling which is 2.4m high. As this ball reaches the ceiling he throws a second ball also at 10m/s. How long after 2nd ball is thrown do they pass

Homework Equations


maybe just too late but I am drawing complete blank. I just worked out the time and velocity of the ball and know the same formulas apply- just how, where do i start?


The Attempt at a Solution

?
 
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Hi Rbraind! :wink:

Do y and t equations for each ball.

You need to find where y and t for one are the same as for the other …

what do you get? :smile:​
 
Rbraind said:

Homework Statement



A juggler throws a ball at 10m/s and it takes 0.24 seconds to reach the ceiling which is 2.4m high. As this ball reaches the ceiling he throws a second ball also at 10m/s. How long after 2nd ball is thrown do they pass

Homework Equations


maybe just too late but I am drawing complete blank. I just worked out the time and velocity of the ball and know the same formulas apply- just how, where do i start?


The Attempt at a Solution

?

I don't think u need an equation for this.
Let x be the time after the 2nd ball has passed
Assume that g=10ms^-1
10x + 10x = 2.4m
20x = 2.4m
x = 0.12 second
 

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