Answer
$|\vec{v_{o}}|=3.07$ m/s
Work Step by Step
We know the following:
$ v_{0y}=0,\qquad$ time of flight = $t=0.495$ s$, \quad\Delta x=1.52$ m.
Since there is no acceleration in the horizontal direction,
$\Delta x=v_{0x}t$
$1.52m=v_{0x}(0.495s)$
$v_{0x}=\displaystyle \frac{1.52m}{0.495s}=3.07$ m/s
which is the speed of the ball at the monent it leaves the table
because
$\vec{v_{o}}=v_{x}\hat{i}+v_{y}\hat{j}$
$|\vec{v_{o}}|=\sqrt{(3.07m/s)^{2}+0^{0}}=3.07$ m/s
