To find the maximum height (the peak) of the projectile, we can use the vertex formula for a quadratic equation in the form \( h(t) = at^2 + bt + c \). Given the equation:
\[ h(t) = -16t^2 + 60t + 0 \]
The coefficients are:
- \( a = -16 \)
- \( b = 60 \)
- \( c = 0 \)
The time at which the maximum height occurs can be calculated using the formula:
\[ t = -\frac{b}{2a} \]
Substituting the values of \( a \) and \( b \):
\[ t = -\frac{60}{2 \times -16} = \frac{60}{32} = \frac{15}{8} = 1.875 \text{ seconds} \]
Now, substitute \( t = 1.875 \) back into the height equation to find the maximum height \( h \):
\[ h(1.875) = -16(1.875)^2 + 60(1.875) + 0 \]
Calculating \( (1.875)^2 \):
\[ (1.875)^2 = 3.515625 \]
Now substituting this value:
\[ h(1.875) = -16(3.515625) + 60(1.875) \]
Calculating \( -16(3.515625) \):
\[ -16(3.515625) = -56.25 \]
Calculating \( 60(1.875) \):
\[ 60(1.875) = 112.5 \]
Now adding these results together:
\[ h(1.875) = -56.25 + 112.5 = 56.25 \]
Thus, the maximum height (range) the ball will reach is:
\[ \boxed{56.25} \text{ feet} \]