Question

h(t) = −16t2 + 80t + 30
find H in this equation

Answer 1

In the equation \( h(t) = -16t^2 + 80t + 30 \), the function \( h(t) \) likely represents the height of an object in motion (such as a projectile) depending on the time \( t \).

If you want to find \( H \), which often denotes the maximum height of a projectile, you can find it by determining the vertex of the parabola represented by the quadratic equation. The maximum height occurs at the vertex of the parabola because the leading coefficient (the coefficient of \( t^2 \)) is negative, indicating that it opens downwards.

The formula for the \( t \)-coordinate of the vertex (maximum point) of a quadratic equation in the form \( at^2 + bt + c \) is given by:

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

For your equation:

- \( a = -16 \)
- \( b = 80 \)

Substituting these values into the formula:

\[
t = -\frac{80}{2 \times -16} = -\frac{80}{-32} = \frac{80}{32} = 2.5
\]

Now, to find the maximum height \( H \), substitute \( t = 2.5 \) back into the height equation \( h(t) \):

\[
H = h(2.5) = -16(2.5)^2 + 80(2.5) + 30
\]

Calculating this step-by-step:

1. \( (2.5)^2 = 6.25 \)
2. \( -16(6.25) = -100 \)
3. \( 80(2.5) = 200 \)

Now, plugging these values back into the equation:

\[
H = -100 + 200 + 30 = 130
\]

Thus, the maximum height \( H \) is:

\[
\boxed{130}
\]