To find the maximum height of the cannonball, we need to find the vertex of the quadratic function $h(t) = -4.9t^2 + 171.5t + 10.8$. The formula for the time at which the vertex occurs (the time when the maximum height is reached) in a quadratic function of the form $ax^2 + bx + c$ is given by:

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

In this case:

• $a = -4.9$
• $b = 171.5$

Substituting these values into the formula:

$t_{\text{max}} = \frac{-171.5}{2(-4.9)} = \frac{-171.5}{-9.8} \approx 17.5 \text{ seconds}$

Now that we know the time at which the maximum height occurs, we substitute $t = 17.5$ into the original height equation $h(t) = -4.9t^2 + 171.5t + 10.8$ to find the maximum height:

$h(17.5) = -4.9(17.5)^2 + 171.5(17.5) + 10.8$ $h(17.5) = -4.9(306.25) + 171.5(17.5) + 10.8$ $h(17.5) = -1500.125 + 3001.25 + 10.8$ $h(17.5) = 1512.925 \, \text{meters}$

Therefore, the maximum height of the cannonball is approximately 1512.9 meters.

Would you like any further details or have any questions?

Related Questions:

1. How would the height change if the coefficient of $t^2$ were positive instead of negative?
2. What is the time when the cannonball hits the ground?
3. How do we interpret the coefficient $171.5$ in the context of this problem?
4. How would the equation change if the cannon were at a different initial height?
5. What would happen to the maximum height if the cannonball was fired with less velocity?

Tip: The vertex formula for quadratics is key in finding maximum or minimum values, especially when the parabola opens downward (maximum) or upward (minimum).