The given formula for the height of the object at time \( t \) is
\[ y(t) = -4t^2 + 10t. \]
This is a quadratic equation in the standard form \( y = at^2 + bt + c \), where \( a = -4 \), \( b = 10 \), and \( c = 0 \). The graph of this equation is a parabola that opens downward (since \( a < 0 \)).
To find the maximum height of the object, we need to determine the vertex of the parabola. The time \( t \) at which the maximum height occurs can be found using the vertex formula:
\[ t = -\frac{b}{2a}. \]
Substituting the values of \( a \) and \( b \):
\[ t = -\frac{10}{2 \cdot -4} = \frac{10}{8} = \frac{5}{4} = 1.25 \text{ seconds}. \]
Next, we will substitute \( t = 1.25 \) back into the height equation to find the maximum height \( y \):
\[ y(1.25) = -4(1.25)^2 + 10(1.25). \]
Calculating \( (1.25)^2 = 1.5625 \):
\[ y(1.25) = -4(1.5625) + 12.5 \]
\[ y(1.25) = -6.25 + 12.5 = 6.25 \text{ feet}. \]
Thus, the maximum height that the object will reach is
\[ \boxed{6.25} \text{ feet}. \]
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