- #1

infraray

- 23

- 0

You should upgrade or use an alternative browser.

- Thread starter infraray
- Start date

- #1

infraray

- 23

- 0

- #2

quantumdude

Staff Emeritus

Science Advisor

Gold Member

- 5,575

- 23

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

- #3

infraray

- 23

- 0

- #4

HallsofIvy

Science Advisor

Homework Helper

- 43,021

- 970

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.

- #5

infraray

- 23

- 0

Share:

- Last Post

- Replies
- 6

- Views
- 209

- Last Post

- Replies
- 6

- Views
- 279

- Last Post

- Replies
- 18

- Views
- 325

- Replies
- 7

- Views
- 241

- Last Post

- Replies
- 4

- Views
- 235

- Replies
- 2

- Views
- 425

- Replies
- 11

- Views
- 312

- Replies
- 4

- Views
- 307

- Replies
- 9

- Views
- 486

- Replies
- 14

- Views
- 536