Question

h(t) = −16t2 + 80t + 30
find h

Answer 1

The function \( h(t) = -16t^2 + 80t + 30 \) is a quadratic function representing the height of an object in motion, where \( h(t) \) is the height at time \( t \). If you would like to find specific values of \( h(t) \) for different values of \( t \), or find particular features of the function (such as the maximum height, roots, or vertex), please specify, and I can help you with that!

Here are steps to determine some important features of the quadratic function:

1. **Finding the Vertex**: The vertex of the parabola can be found using the formula:
\[
t = -\frac{b}{2a}
\]
where \( a = -16 \) and \( b = 80 \).

- Calculate \( t \):
\[
t = -\frac{80}{2 \cdot -16} = \frac{80}{32} = 2.5
\]

- Substitute \( t = 2.5 \) back into \( h(t) \) to find the maximum height:
\[
h(2.5) = -16(2.5)^2 + 80(2.5) + 30
\]
\[
h(2.5) = -16(6.25) + 200 + 30
\]
\[
h(2.5) = -100 + 200 + 30 = 130
\]

So, the maximum height is \( 130 \) units at \( t = 2.5 \) seconds.

2. **Finding the Roots**: The roots can be found using the quadratic formula:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For our function:
- \( a = -16 \)
- \( b = 80 \)
- \( c = 30 \)

Calculate the discriminant:
\[
b^2 - 4ac = 80^2 - 4(-16)(30) = 6400 + 1920 = 8320
\]

Now applying the quadratic formula:
\[
t = \frac{-80 \pm \sqrt{8320}}{2 \cdot -16}
\]
\[
t = \frac{-80 \pm \sqrt{8320}}{-32}
\]

Calculate \( \sqrt{8320} \) (approximately 91.1):
\[
t = \frac{-80 \pm 91.1}{-32}
\]

This gives two values for \( t \):
- \( t_1 \approx \frac{11.1}{-32} \approx -0.35 \) (not physically meaningful in this context)
- \( t_2 \approx \frac{-171.1}{-32} \approx 5.34 \)

Thus, the object hits the ground again at approximately \( t = 5.34 \) seconds.

Let me know if you need further analysis or help!