Question

A steel ball hangs in a 147cm thread which is dragged right and then let go. Velocityh of the ball is 2,44m/s when it goes through the middle position.

I don't have any mass given up.

How many degrees to the side had the ball been dragged?

Answer 1

ke of ball at bottom = (1/2) m (2.44)^2
= 2.98 m

Pe at top = m g h = m (9.81)(h)
so
9.81 h m = 2.98 m
h = .303 meters

h = 1.47 (1 - cos theta)
.303/1.47 = 1 - cos T
.206 = 1 - cos T
-cos T = -.794
T = 37.4 degrees

Answer 2

thank you very much ! i got it half the way but couldn't figure out how to finish it .

Answer 3

You are welcome :)