What is the Vertical Separation of Steel Balls in a Steady Fall?

AI Thread Summary
The discussion focuses on calculating the vertical separation of two steel balls falling at a steady rate of two per second. The lower ball has fallen 3 meters, taking approximately 0.782 seconds to reach that distance. The upper ball, dropped half a second earlier, is still in motion, and its position is determined by the time difference. The correct approach involves understanding the timing of the drops and applying kinematics equations to find the vertical separation. Ultimately, the participants clarify the timing and calculations needed to solve the problem accurately.
Larrytsai
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Homework Statement


Small Steel balls fall from rest through the opening at A at the steady rate of two per second. Find the vertical separation h of two consecutive balls when the lower one has dropped 3 meters.


Homework Equations


all kinematics equations


The Attempt at a Solution


I found the time it took for 1 ball to reach 3m which was 0.782 seconds, lost from here
 
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Ok, so the lower one was dropped 0.728s ago and is 3m down. Then the upper one was dropped half a second earlier. Where is it?
 
so then, i go 0.782/2 is the time for the second ball?
 
Larrytsai said:
so then, i go 0.782/2 is the time for the second ball?

Nooo! 0.782s-(1/2)s. It was dropped 1/2 second earlier. They are falling at a rate of 2/second. Think about it.
 
ohhh that makes sense, thanx
 

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