To find the maximum height of the object thrown into the air, we can analyze the quadratic function given by the equation:
\[ y = -4t^2 + 10t \]
This is a parabola that opens downwards (due to the negative coefficient of \(t^2\)). The maximum height occurs at the vertex of the parabola.
The time \(t\) at which the maximum occurs can be found using the formula:
\[ t = -\frac{b}{2a} \]
where \(a\) and \(b\) are the coefficients from the standard quadratic form \(y = at^2 + bt + c\). In this case, \(a = -4\) and \(b = 10\).
Plugging in these values:
\[ t = -\frac{10}{2(-4)} = \frac{10}{8} = 1.25 \text{ seconds} \]
Next, we substitute \(t = 1.25\) back into the original equation to find the maximum height:
\[ y = -4(1.25)^2 + 10(1.25) \] \[ y = -4(1.5625) + 12.5 \] \[ y = -6.25 + 12.5 = 6.25 \text{ feet} \]
Therefore, the maximum height the object will reach is 6.25 feet.