Answer
(a) $v_0 = 17.9~m/s$
(b) $y = 16.3~m$
Work Step by Step
Let $t$ be the time it takes for the entertainer to move to the table. During this time, the ball will go up to the maximum height.
$t = \frac{d}{v} = \frac{5.50 ~m}{3.00 ~m/s} = 1.83 ~s$
We can use this time $t$ to find the minimum initial speed of the ball.
$v = v_0 + at$
$v_0 = 0 -at = -at = -(-9.80~m/s^2)(1.83~s)$
$v_0 = (9.80 ~m/s^2)(1.83~s) = 17.9~m/s$
(b) $y = \frac{v_0^2}{2g} = \frac{(17.9~m/s)^2}{(2)(9.80~m/s^2)}$
$y = 16.3~m$
