if a rock is falling at 10 m/s and was dropped near the surface of a planet where g = 20 m/s^2 by how much would its speed change each second
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if a rock is falling at 10 m/s and was dropped near the surface of a planet where g = 20 m/s^2 by how much would its speed change each second
The answer is in the question :)
Capuchin is correct by saying the answer is in the question. Think about it. No matter what the object weight is it will fall down because of gravity. So if the gravity is 9.8 m/s^2 and you jumped of an airplane you would be falling at a rate of 9.8 m/s^2 increase until you reach your terminal velocity. So if the acceleration is 20 m/s^2 than it will be same concept :)
Well, I'm assuming there is no air resistance. cool_dude, that isn't how terminal velocity works, your acceleration decreases until it reaches 0, it doesn't just hit terminal velocity and stop/
Even if there was air resistance it is negligible or at least that is what we do in our physics class. Yes I didn't word it correctly it doesn't just hit terminal velocity thanks for rewording it.
Yes, for a rock it is negligible, due to high density/rigidity :)
Can't talk about terminal velocity and keep air resistance negligible. Excepting the case where velocity approaches c, there's no terminal velocity in a vacuum, and I think relativity is a bit beyond the scope of the question :) (although it adds a fun dimension to the answer if you want to play with it)Quote:
Originally Posted by cool_dude
Terminal velocity is the point at which the acceleration due to gravity (I suppose you could use it for any motive force, but the term is (from what I understand, anyway) limited to falling-body issues) is offset by the drag the object experiences, which (generally)increases with velocity.
I think you can use terminal velocity an any direction, but of course it's difficult to get an attractive force to act so strongly and for such a long time that the terminal velocity is reached in a direction other than toward the center of the earth :).
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