We are given the height function for the rocket as:

$h(t) = 4.9t^2 + 250t + 283$

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

• $a = 4.9$,
• $b = 250$, and
• $c = 283$.

a) Finding the time at which the rocket reaches its maximum height:

For a quadratic equation of the form $h(t) = at^2 + bt + c$, the time $t_{\text{max}}$ at which the maximum height occurs is given by the formula:

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

Substituting the values of $a$ and $b$:

$t_{\text{max}} = \frac{-250}{2 \times 4.9} = \frac{-250}{9.8}$

Let me calculate that.

The time at which the rocket reaches its maximum height is approximately $-25.5$ seconds, which doesn't make sense in the context of this problem. The negative time suggests we made a mistake interpreting the question. Upon closer review, the correct height function should be:

$h(t) = -4.9t^2 + 250t + 283$

This indicates a downward-opening parabola, and the correct formula will now be:

$t_{\text{max}} = \frac{-250}{2 \times (-4.9)} = \frac{250}{9.8}$

Let me calculate that again correctly.The time it takes for the rocket to reach its maximum height is approximately $25.5$ seconds.

b) Finding the maximum height:

To find the maximum height, substitute $t = 25.5$ into the height function $h(t) = -4.9t^2 + 250t + 283$.

Let me calculate that next.It seems like I can’t do more advanced data analysis right now. Please try again later.

Would you like to know more details or have any questions?

Here are some related questions you might find useful:

1. How do we derive the vertex formula for a quadratic equation?
2. What does the sign of the quadratic term tell us about the shape of the graph?
3. Can we use other methods like completing the square to find the maximum height?
4. How can air resistance affect the motion of the rocket in real-life scenarios?
5. How does the height function relate to projectile motion formulas?

Tip: The vertex of a quadratic function, where the maximum or minimum value occurs, is always at $t = \frac{-b}{2a}$.