Question

An object is thrown the air y = -4t^2 +10t. What is the maximum height of the object in feet

Answer 1

The equation given for the height \( y \) of the object in terms of time \( t \) is:

\[
y = -4t^2 + 10t
\]

This is a quadratic equation in the form:

\[
y = at^2 + bt + c
\]

where \( a = -4 \), \( b = 10 \), and \( c = 0 \).

To find the maximum height, we can use the formula for the vertex of a parabola, given by \( t = -\frac{b}{2a} \). Substituting in our values for \( a \) and \( b \):

\[
t = -\frac{10}{2(-4)} = \frac{10}{8} = 1.25 \text{ seconds}
\]

Next, we substitute \( t = 1.25 \) back into the equation to find the maximum height:

\[
y = -4(1.25)^2 + 10(1.25)
\]

Calculating \( (1.25)^2 \):

\[
(1.25)^2 = 1.5625
\]

Now substitute:

\[
y = -4(1.5625) + 10(1.25)
\]

Calculating:

\[
y = -6.25 + 12.5 = 6.25
\]

Thus, the maximum height of the object is:

\[
\boxed{6.25} \text{ feet}
\]