Answer
The initial launch velocity is 4.85 m/s and the time in the air is 0.99 seconds.
Work Step by Step
(a) $v_0^2 = 2ay$
$v_0 = \sqrt{2ay} = \sqrt{(2)(9.80)(1.20)} = 4.85~m/s$
(b) The time $t$ it takes to reach maximum height is:
$t = \frac{v_0}{g} = \frac{4.85}{9.80} = 0.495 ~s$
The total time in the air is $2t$ which is 0.99 seconds.
