#### Answer

The maximum height reached is $\frac{221}{16}$ft or 13.8 ft.
The range is $0\leq h \leq \frac{221}{16}$

#### Work Step by Step

$h = -16t^{2} + 30t + 6$
The standard form for a quadratic equation is
$y = ax^{2} + bx + c$ So a= -16, b= 30, and c= 6
Axis of symmetry: In this question the Axis of symmetry gives us the time to reach the maximum height.
The formula for axis of symmetry is
$x= \frac{-b}{2a}$
$x= \frac{-(30)}{2(-16)}$
$x= \frac{-30}{-32}$
$x= \frac{15}{16}$
Vertex: In this question the vertex gives us the maximum height of the ball.
Plug in the x value of the axis of symmetry to find the y value of the vertex.
$h = -16t^{2} + 30t + 6$
$h = -16(\frac{15}{16})^{2} + 30(\frac{15}{16}) + 6$
y= $\frac{221}{16}$
The vertex is ($\frac{15}{16}$, $\frac{221}{16}$)
Since the ball starts at ground level of zero and goes up to $\frac{221}{16}$ the range is $0\leq h \leq \frac{221}{16}$