Asmall lead ball, attached to a 1.25-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 1.5 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?

Explanation:

Radius of the circular path, r = 1.25 m

Angular velocity of the ball,

It is placed 1.5 meters above the ground. Using the conservation of energy to find the height of the ball.

Since,

h = 28.29 meter

So, the ball will rise to a maximum height of 1.5 m + 28.29 m = 29.796 meters. Hence, this is the required solution.

a straight line sloped

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach