- #1

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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?

- Thread starter chaotixmonjuish
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- #1

- 287

- 0

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?

- #2

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- #3

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here are some hints which maybe able to help you

[tex]s = ut + \frac{1}{2}at^2 [/tex]

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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