Answer:
The y-intercept is when $x=0$ so $y=32\frac{ft}{s}*0-\frac{1}{2}*9.8*0^2=0$, so the ball starts on the ground. The maximum height is when $v=0$, so $0=32\frac{ft}{s}-9.8t$ and solving for $t$ we get $t=\frac{32}{9.8}=3.265s$, and plugging this back in for $y$ we get $y=32\frac{ft}{s}*3.265s-\frac{1}{2}*9.8*3.265^2=156.205 ft$. So the ball will be at its maximum height of 156.205 ft at 3.265 seconds.
Explanation:
The $y$-intercept is $-1$. The maximum height of the baseball is when $y=0$; that is when $x=\ln(4)$. Here is a sketch of the path.
From the $y$-intercept, we see that the baseball starts at height $y=-1$. It reaches a maximum height at $x=\ln(4)$; that is, $y=0$. From there, it falls and reaches the ground at $x=2\ln(4)$; that is, $y=-1$ again.