Answer
(a) $0.27rev$
(b) It does not depend on her initial speed.
Work Step by Step
As we know that
$y=y_{\circ}+v_{\circ}t-\frac{1}{2}gt^2$
$\implies 0=y_{\circ}+0-\frac{1}{2}gt^2$
This can be rearranged as:
$t=\sqrt{\frac{2y_{\circ}}{g}}$
The number of revolutions can be determined as
$\Delta \theta=\omega_{avg}\Delta t$
$\implies \Delta \theta=\omega_{avg}\sqrt{\frac{2y_{\circ}}{g}}$
We plug in the known values to obtain:
$\Delta theta=(\frac{2.2rad}{s})(\frac{1rev}{2\pi rad})(\sqrt{\frac{2(3.0)}{9.81}}=0.27rev$
(b) We know that the answer to part (a) does not depend on the initial speed. The initial speed only helps us to determine that how far she goes horizontally before she hits the water.
