Answer
$t=0.495$ s
Work Step by Step
(a)
We observe the motion of the ball component by component.
Vertically, the instant it leaves the table, it has vertical initial velocity of zero,
and proceeds down under constant acceleration. We can use Table 2-1.
The ball starts at 1.20m and ends at 0 m (we take up as +y)
$\displaystyle \Delta y=v_{oy}t+\frac{1}{2}at^{2} $
$-1.20m=\displaystyle \frac{1}{2}(-9.80m/s^{2})t^{2}$
$t^{2}=\displaystyle \frac{1.20m}{9.80m/s^{2}}$
$t=0.495$ s
