The problem involves finding the highest point (maximum height) that the object reaches after being launched. The height of the object as a function of time is given by the quadratic equation:

$h(t) = -16t^2 + 160t + 624$

This is a quadratic equation in the form $h(t) = at^2 + bt + c$, where:

• $a = -16$
• $b = 160$
• $c = 624$

Step 1: Find the time at which the object reaches its maximum height.

The maximum (or minimum) of a quadratic function $h(t) = at^2 + bt + c$ occurs at the time $t$ given by the formula:

$t_{\text{max}} = \frac{-b}{2a}$

Substitute the values for $a$ and $b$:

$t_{\text{max}} = \frac{-160}{2(-16)} = \frac{-160}{-32} = 5 \, \text{seconds}$

Step 2: Find the maximum height by substituting $t = 5$ into the height equation.

$h(5) = -16(5)^2 + 160(5) + 624$

First, calculate each term:

$-16(5)^2 = -16(25) = -400$ $160(5) = 800$

Now substitute:

$h(5) = -400 + 800 + 624 = 1024 \, \text{feet}$

Conclusion:

The highest point that the object reaches is 1024 feet.

Would you like more details or have any questions?

Here are some related questions you can explore:

1. How do you determine the time at which an object reaches the highest or lowest point in a quadratic function?
2. What would happen if the object's initial velocity were different, and how would that change the height reached?
3. How can we calculate the time it takes for the object to return to the ground (when $h(t) = 0$)?
4. How does the negative coefficient in front of $t^2$ affect the graph of the height function?
5. Can you derive the formula for the maximum height of any object launched vertically, using a similar equation?

Tip: Quadratic equations with negative coefficients for $t^2$ always have a downward-facing parabola, meaning they reach a maximum at the vertex.