Answer
45 m. 6 s.
Work Step by Step
The ball's initial velocity is 30 m/ s and it loses 10 m/s each second. It spends 3 seconds going up until v = 0, and the same time coming down. It’s in the air for a total of 6 seconds.
At the top of its flight, the ball is momentarily at rest. After that, the distance it drops in 3 s can be calculated using an equation on page 48.
$$d=\frac{1}{2}gt^{2}=\frac{1}{2}\times10\frac{m}{s^{2}}\times(3s)^{2}=45 m$$
