To find the maximum height of the object described by the equation \( y = -4t^2 + 10t \), we can use the vertex formula for a quadratic equation of the form \( y = ax^2 + bx + c \).
For the equation \( y = -4t^2 + 10t \):
- \( a = -4 \)
- \( b = 10 \)
The time \( t \) at which the maximum height occurs can be found using the formula:
\[ t = -\frac{b}{2a} \]
Plugging in the values of \( a \) and \( b \):
\[ t = -\frac{10}{2(-4)} = \frac{10}{8} = 1.25 \text{ seconds} \]
Now, we substitute \( t = 1.25 \) back into the original equation to find the maximum height \( y \):
\[ y = -4(1.25)^2 + 10(1.25) \]
Calculating \( (1.25)^2 \):
\[ (1.25)^2 = 1.5625 \]
Now substituting:
\[ y = -4(1.5625) + 10(1.25) \]
Calculating further:
\[ y = -6.25 + 12.5 = 6.25 \text{ feet} \]
Thus, the maximum height of the object is 6.25 feet.