To find the maximum height (or the range) of the ball using the equation \( h = -16t^2 + 60t + 0 \), we can use the vertex formula for a quadratic equation, which gives the time at which the maximum height occurs. For a quadratic equation in the form \( h = at^2 + bt + c \), the \( t \) coordinate of the vertex is given by:
\[ t = -\frac{b}{2a} \]
In this case, \( a = -16 \) and \( b = 60 \).
Plugging in these values:
\[ t = -\frac{60}{2 \times -16} = -\frac{60}{-32} = \frac{60}{32} = \frac{15}{8} = 1.875 \text{ seconds} \]
Now, we substitute this \( t \) value back into the height equation to find the maximum height:
\[ h = -16(1.875)^2 + 60(1.875) + 0 \]
Calculating \( (1.875)^2 \):
\[ (1.875)^2 = 3.515625 \]
Now substituting back into the equation:
\[ h = -16(3.515625) + 60(1.875) \]
Calculating each part:
\[ -16(3.515625) = -56.25 \] \[ 60(1.875) = 112.5 \]
Now adding these results:
\[ h = -56.25 + 112.5 = 56.25 \text{ feet} \]
Thus, the maximum height (or range) the ball will reach is:
\[ \boxed{56.25} \text{ feet} \]