Explanation:
Radius of the circular path, r = 1.25 m
Angular velocity of the ball, [tex]\omega=3\ rev/s=18.84\ rad/s[/tex]
It is placed 1.5 meters above the ground. Using the conservation of energy to find the height of the ball.
[tex]mgh=\dfrac{1}{2}mv^2[/tex]
[tex]h=\dfrac{v^2}{2g}[/tex]
Since, [tex]v=r\times \omega[/tex]
[tex]h=\dfrac{(r\omega)^2}{2g}[/tex]
[tex]h=\dfrac{(1.25\times 18.84)^2}{2\times 9.8}[/tex]
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.
