Question

Some fireworks are fired vertically into the air from the ground at an initial speed of 80 ft./s equation for this objects height H at time T seconds after lunch is h= -16t^2 +80t

Answer 1

The equation given for the height \( h \) of the fireworks over time \( t \) seconds is:

\[
h(t) = -16t^2 + 80t
\]

This equation represents a quadratic function where:

- The term \(-16t^2\) represents the effect of gravity (in feet, since \( 16 \) is half the acceleration due to gravity, which is approximately \( 32 \, \text{ft/s}^2 \)).
- The term \( 80t \) represents the initial upward velocity of the fireworks.

### Key Features of the Equation

1. **Initial Height**: At \( t = 0 \):
\[
h(0) = -16(0)^2 + 80(0) = 0
\]
So, the fireworks start at ground level.

2. **Maximum Height**: To find the maximum height, we can find the vertex of the parabola. The time \( t \) at which the maximum height occurs can be found using the formula:
\[
t = -\frac{b}{2a}
\]
where \( a = -16 \) and \( b = 80 \):
\[
t = -\frac{80}{2(-16)} = \frac{80}{32} = 2.5 \, \text{seconds}
\]

Now, substitute \( t = 2.5 \) back into the height equation to find the maximum height:
\[
h(2.5) = -16(2.5)^2 + 80(2.5)
\]
\[
h(2.5) = -16(6.25) + 80(2.5)
\]
\[
h(2.5) = -100 + 200 = 100 \, \text{feet}
\]

3. **Time to Reach the Ground**: To find when the fireworks hit the ground, set \( h(t) = 0 \):
\[
-16t^2 + 80t = 0
\]
Factor out \( t \):
\[
t(-16t + 80) = 0
\]
This gives us \( t = 0 \) (when launched) and \( -16t + 80 = 0 \):
\[
-16t = -80 \implies t = 5 \, \text{seconds}
\]

### Summary

- The maximum height of the fireworks is **100 feet**, occurring at **2.5 seconds** after launch.
- The fireworks return to the ground after **5 seconds**.

If you have further questions or need additional information, feel free to ask!