A tetherball apparatus consists of a ball attached to a 1.50 m rope that is atta

Question

A tetherball apparatus consists of a ball attached to a 1.50 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 4.71 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?
....
If you start over and hit the ball harder, such that it moves with speed 12.1 m/s at a height of 2.55 m, what would the centripetal acceleration of the ball have to be (as a multiple of g)?


A tetherball apparatus consists of a ball attached to a 1.50 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 4.71 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?
....
If you start over and hit the ball harder, such that it moves with speed 12.1 m/s at a height of 2.55 m, what would the centripetal acceleration of the ball have to be (as a multiple of g)?


A tetherball apparatus consists of a ball attached to a 1.50 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 4.71 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? ....
If you start over and hit the ball harder, such that it moves with speed 12.1 m/s at a height of 2.55 m, what would the centripetal acceleration of the ball have to be (as a multiple of g)?


....
If you start over and hit the ball harder, such that it moves with speed 12.1 m/s at a height of 2.55 m, what would the centripetal acceleration of the ball have to be (as a multiple of g)?

If you start over and hit the ball harder, such that it moves with speed 12.1 m/s at a height of 2.55 m, what would the centripetal acceleration of the ball have to be (as a multiple of g)?
If you start over and hit the ball harder, such that it moves with speed 12.1 m/s at a height of 2.55 m, what would the centripetal acceleration of the ball have to be (as a multiple of g)?

Explanation / Answer

inclination with vertical = theta

tan theta = V^2/(r*g)

theta = 60.11 34

vertical height from ground = 2.7 - (1.50cos60.1134) = 1.952 m

centripetal acceleration = V^2/rg doesnot depend on height

= 9.959 m/s^2 = 1.016* g

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