Answer
The maximum height the ball would reach is $41.5m$
Work Step by Step
1) Find the release velocity.
The ball is whirled at a rate of 3 revolutions per second, so its period is $T=1/3s$. The radius of the whirling circle is $r=1.5m$
$$v=\frac{2\pi r}{T}=28.27m/s$$
The speed in a uniform circular motion is constant, and so is the release velocity. This means the release velocity is $28.27m/s$
2) After the ball is released, it travels straight upward, so its KE decreases while its PE increases by the same amount, following the principle of energy conservation. Therefore, $$\Delta PE = -\Delta KE$$ $$mg\Delta h=-\frac{1}{2}m\Delta v^2$$ $$g\Delta h=-\frac{1}{2}(v_f^2-v_0^2) (1)$$
$g=9.8m/s^2$, the initial release speed $v_0=28.27m/s$, and the maximum height is when $v_f=0$ .
Using these, we can find the change in height: $$9.8\Delta h=399.6$$ $$\Delta h=40.78m$$
The maximum height the ball would reach is $0.75+40.78=41.53m$
