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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?

Answer :

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

[tex]velocity = \dfrac{distance}{time}[/tex]

circumference of the circle = 2 π r

                                             = 2 x π x 1.7

                                             = 10.68 m

[tex]velocity = \dfrac{3 \times 10.68}{1}[/tex]

v = 32.04 m/s

using conservation of energy

[tex]\dfrac{1}{2}mv^2 + mgh = mgH[/tex]

[tex]\dfrac{1}{2}v^2 + gh = gH[/tex]

[tex]\dfrac{1}{2}32.04^2 + 9.8\times 2 = 9.8\times H[/tex]

[tex]532.88 = 9.8\times H[/tex]

H = 54.37

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

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