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baherus [9]
3 years ago
14

A small lead ball, attached to a 1.70-m rope, is being whirled in a circle that lies in the vertical plane. The ball is whirled

at a constant rate of three revolutions per second and is released on the upward part of the circular motion when it is 2.0 m above the ground. The ball travels straight upward. In the absence of air resistance, to what maximum height above the ground does the ball rise?
Physics
1 answer:
Otrada [13]3 years ago
4 0

Answer:

H = 54.37

Explanation:

given,

lead ball attached at = 1.70 m

rate of revolution = 3 revolution/sec

height above the ground = 2 m

velocity = \dfrac{distance}{time}

circumference of the circle = 2 π r

                                             = 2 x π x 1.7

                                             = 10.68 m

velocity = \dfrac{3 \times 10.68}{1}

v = 32.04 m/s

using conservation of energy

\dfrac{1}{2}mv^2 + mgh = mgH

\dfrac{1}{2}v^2 + gh = gH

\dfrac{1}{2}32.04^2 + 9.8\times 2 = 9.8\times H

532.88 = 9.8\times H

H = 54.37

the maximum height reached by the ball is equal to H = 54.37

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