Asked by Abigail
An astronaut on the moon drops a rock straight downward from a height of 0.95m. If the acceleration due to gravity on the moon is 1.62m/s, what is the speed of the rock just before it lands?
Answers
Answered by
Reiny
Just give me the answer??
That's not how this works. Where do you have a problem?
Btw, the acceleration should be negative, or else will just keep flying off the moon and
acceleration would be -1.62m/s^2 , not m/s
That's not how this works. Where do you have a problem?
Btw, the acceleration should be negative, or else will just keep flying off the moon and
acceleration would be -1.62m/s^2 , not m/s
Answered by
oobleck
v = √(2as)
How does this work? well, recall that
s = 1/2 at^2 so, t = √(2s/a)
But at the same time, v = at = a√(2s/a) = √(2as)
So just plug in your numbers, and take the time to learn a few elementary but very useful formulas.
How does this work? well, recall that
s = 1/2 at^2 so, t = √(2s/a)
But at the same time, v = at = a√(2s/a) = √(2as)
So just plug in your numbers, and take the time to learn a few elementary but very useful formulas.
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