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 2.50 m tall pole. The ball is struck such that it moves horizontally around the pole. Immediately after the ball is struck, it moves at 3.34 m/s, and experiences a centripetal acceleration of 0.842g (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

if rope is making an angle with pole than ,

on ball , in vertical,

Tcos = mg

in horizontal, Tsin = mv2 / lsin

centripiatl acc. given mv2 / lsin = 0.842g

m3.342 / 2.10sin = 0.842 x 9.81

m/sin = 1.56

Tsin / Tcos = mv2 /mlgsin

tan = 3.342 / 2.10 x 9.81 x sin

sin2 = 0.542cos

1 - cos2 = 0.542cos

cos2 + 0.542cos - 1 = 0

cos = 0.765

= 40.09 degrees

h = 2.50 - 2.10cos40.09 = 0.89 m

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