A tetherball apparatus consists of a ball attached to a 2.10 m rope that is atta

Question

A tetherball apparatus consists of a ball attached to a 2.10 m rope that is attached in turn to the top of a 2.70 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?

Explanation / Answer

a = v^2/r

r = v^2/a = 5.57^2/(1.74*9.81)=1.818 m

so dy = sqrt(R^2 - r^2)

so off the ground = 2.7 - sqrt(2.1^2 - 1.818^2)=1.649 m

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