## University Physics with Modern Physics (14th Edition)

(a) $v_0 = 17.9~m/s$ (b) $y = 16.3~m$
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) $y = \frac{v_0^2}{2g} = \frac{(17.9~m/s)^2}{(2)(9.80~m/s^2)}$ $y = 16.3~m$