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- Thread starter infraray
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Tom Mattson

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Originally posted by infraray

I assume I am to figure t first to then find v, where t= sqrt(2y/g), which yields approx. 0.64s.

That equation is for free fall, not for circular motion.

The acceleration of a particle undergoing uniform circular motion is

a

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HallsofIvy

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You know that when y= 0, x= 10 m. You should be able to solve the equations (-g/2)t^2+ 2= 0 for t (yes, t= sqrt(2y/g) which, since y= 2, is t= 2/sqrt(g)) and then v t= 10 for v. THAT gives you the speed of the stone. Knowing that you can use the formulas for circular motion (particularly a= v^2/R) to find the acceleration, a.

By the way, you say that "The answer is supposed to be 160 m/s."

Of course, that's impossible: I'm sure you mean 160 m/s^2.

Actually, I get 163 1/3 meters per second squared.

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