An elevator moving up and a bolt falling down

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SUMMARY

The discussion centers on calculating the height of an elevator when a bolt falls from it while moving upward at 6.38 m/s. The bolt takes 3.85 seconds to reach the bottom of the elevator shaft. The correct approach involves using the equation of motion: s = ut + ½at², where the initial velocity (u) is -6.38 m/s (due to the opposing direction of gravity) and acceleration (a) is -9.8 m/s². By substituting the values into the equation, the height of the elevator can be accurately determined.

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A bolt comes loose from the bottom of an elevator that is moving upward at a speed of 6.38 m/s. The bolt reaches the bottom of the elevator shaft in 3.85 s. How high up was the elevator when the bolt came loose?

I integrated 6.38 to get the position equation and plugged in 3.85. However, I'm not really sue that's right at all. Could someone show me what type of equation I should be using?
 
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Should I get the height of the elevator at 3.85 seconds. Then use 9.8m/s and create a position equation for the falling bolt.
 
firstly you have posted this in the wrong forum

here are some hints which maybe able to help you

s = ut + \frac{1}{2}at^2

the velocity of the elevator is acting in the opposite direction to gravity, therefore it is a negative velocity in respect to gravity so gravity will be 9.8m/s/s and the elevator velocity will be -6.8m/s

you have the value of t so use the equation above, where s is the displacement, i.e., how high up the elevator was in the shaft
 

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