To find the maximum height of the projectile, we need to determine the vertex of the quadratic function $h(t) = -16t^2 + 64t$.

Step 1: Identify the coefficients

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

• $a = -16$
• $b = 64$
• $c = 0$ (since there is no constant term).

Step 2: Find the time at which the maximum height occurs

The time at which the maximum height occurs can be found using the formula for the vertex of a parabola: $t = -\frac{b}{2a}$ Substituting the values of $a$ and $b$: $t = -\frac{64}{2(-16)} = \frac{64}{32} = 2 \text{ seconds}$

Step 3: Calculate the maximum height

Now, substitute $t = 2$ seconds back into the height function $h(t)$ to find the maximum height: $h(2) = -16(2)^2 + 64(2)$ $h(2) = -16(4) + 128 = -64 + 128 = 64 \text{ feet}$

Conclusion

The maximum height of the projectile is 64 feet.

Would you like further details or have any questions?

Here are 5 related questions you might be interested in:

1. How do we derive the formula for the vertex of a parabola?
2. What would the height be at $t = 1$ second?
3. How long will it take for the projectile to hit the ground?
4. What is the significance of the coefficient $a$ in the quadratic function?
5. How would the problem change if the initial velocity were different?

Tip: The vertex form of a quadratic function provides an easy way to identify the maximum or minimum value of the function.