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my question is: is it possible to find the acceleration due to gravity? (keep in mind this is not taken from earth.)

- Thread starter ACLerok
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- #1

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my question is: is it possible to find the acceleration due to gravity? (keep in mind this is not taken from earth.)

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jamesrc

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[tex] a = \frac{\Delta v}{\Delta t} [/tex]

Here, v

- #3

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Knowing the original velocity and Time, acceleration is easy to calculate.

lol, you beat me to it James!

- #4

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so is the acceleration 28/9 m/s^2?

- #5

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I missed something here. The original question stated: *a 2.50-kg stone thrown upward from the ground at 14. m/s returns to the ground in 9. s*

This means that it took 9 seconds from the time it left the ground until it fell back down. that's 4.5 seconds each way. When the stone reaches it's maximum height, it will have a velocity of 0 m/s. We know that it will impact the ground at a V of 14 m/s and it will take 4.5 seconds to go from 0 m/s to 14 m/s. So...

[tex] a = \frac{V}{t} = \frac{14 m/s}{4.5 s} = 3.11 m/s^2[/tex]

This means that it took 9 seconds from the time it left the ground until it fell back down. that's 4.5 seconds each way. When the stone reaches it's maximum height, it will have a velocity of 0 m/s. We know that it will impact the ground at a V of 14 m/s and it will take 4.5 seconds to go from 0 m/s to 14 m/s. So...

[tex] a = \frac{V}{t} = \frac{14 m/s}{4.5 s} = 3.11 m/s^2[/tex]

Last edited:

- #6

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Actually that works out as well since the round trip took nine seconds and +14 m/s - -14 m/s = 28 m/s. So 28/9 is correct as well.Originally posted by ACLerok

so is the acceleration 28/9 m/s^2?

ugh, it's to late for this.

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Not to nitpick, but isn't "same speed" more appropriate than "same velocity"?Originally posted by Jimmy

Yes. The stone will have the same velocity when it hits the ground as it did when it originally left.

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Please, nitpick. I goofed. Yes, same speed would be more appropriate.

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