Physics
posted by Leah
A tetherball apparatus consists of a ball attached to a 2.10 m rope that is attached in turn to the top of a 3.10 m tall pole. The ball is struck such that it moves horizontally around the pole. Immediately after the ball is struck, it moves at 5.57 m/s, and experiences a centripetal acceleration of 1.74g (where g is the acceleration due to gravity of a falling object near Earth\'s surface). If the resulting path of the ball is approximately a circle, what is the height h of the ball above the ground?
And if you start over and hit the ball harder, such that it moves with speed 16.9 m/s at a height of 2.95 m, what would the centripetal acceleration of the ball have to be (as a multiple of g)?

bobpursley
centruoetal=v^2/r=1.74g
but theta=arctan(1.74g*1/g)=arctan(1.74)
height above ground=3.102.10*cosTheta
sketch the diagram, theta is the angle between the rope and the pole. 
Leah
I figured out the first part but I'm having trouble with the second part
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