Question

From 4 ft above the ground Maya throws a ball upward with a velocity of 18 ft per second the height h(t) of the ball t seconds after Maya throws the ball is given by h(t)= -16t^2 +18t+4 find the maximum height reached by the ball and the time that this height is reached.

Answer 1

To find the maximum height reached by the ball and the time at which it is reached, we start with the height function given by:

\[
h(t) = -16t^2 + 18t + 4
\]

This is a quadratic function in the standard form \( h(t) = at^2 + bt + c \), where:
- \( a = -16 \)
- \( b = 18 \)
- \( c = 4 \)

Since \( a < 0 \), this parabola opens downward, which means it has a maximum point.

The time at which the maximum height occurs can be found using the formula:

\[
t = -\frac{b}{2a}
\]

Substituting in the values of \( a \) and \( b \):

\[
t = -\frac{18}{2 \times -16} = -\frac{18}{-32} = \frac{18}{32} = \frac{9}{16} \text{ seconds}
\]

Next, we need to find the maximum height by substituting this value of \( t \) back into the height function \( h(t) \):

\[
h\left(\frac{9}{16}\right) = -16\left(\frac{9}{16}\right)^2 + 18\left(\frac{9}{16}\right) + 4
\]

Calculating each term:

1. Calculate \( \left(\frac{9}{16}\right)^2 \):

\[
\left(\frac{9}{16}\right)^2 = \frac{81}{256}
\]

2. Substitute into the height function:

\[
h\left(\frac{9}{16}\right) = -16 \cdot \frac{81}{256} + 18 \cdot \frac{9}{16} + 4
\]

\[
= -\frac{16 \cdot 81}{256} + \frac{18 \cdot 9}{16} + 4
\]

\[
= -\frac{81}{16} + \frac{162}{16} + 4
\]

\[
= -\frac{81}{16} + \frac{162}{16} + \frac{64}{16}
\]

Now simplifying this:

\[
= \frac{-81 + 162 + 64}{16} = \frac{145}{16}
\]

Thus, the maximum height is:

\[
\frac{145}{16} \text{ feet} \approx 9.0625 \text{ feet}
\]

### Summary:

- The time at which the maximum height is reached: \( \frac{9}{16} \text{ seconds} \)
- The maximum height reached by the ball: \( \frac{145}{16} \text{ feet} \approx 9.0625 \text{ feet} \)