Let's solve the given problem step by step. The height of the diver is modeled by the quadratic function:

$h(t) = 5t - 10t^2 + 10$

Part (a): How high is the springboard above the water?

To find how high the springboard is above the water, evaluate $h(t)$ when $t = 0$ (the time at which the diver starts the dive):

$h(0) = 5(0) - 10(0)^2 + 10 = 10$

Answer: The springboard is 10 meters above the water.

Part (b): Use the model to find the time at which the diver hits the water.

The diver hits the water when the height, $h(t)$, is zero. Set the height function equal to zero and solve for $t$:

$h(t) = 0 = 5t - 10t^2 + 10$ This simplifies to: $-10t^2 + 5t + 10 = 0$

We can solve this quadratic equation using the quadratic formula: $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ For the equation $-10t^2 + 5t + 10 = 0$, the coefficients are: $a = -10, \quad b = 5, \quad c = 10$

Substitute these into the quadratic formula: $t = \frac{-5 \pm \sqrt{5^2 - 4(-10)(10)}}{2(-10)} = \frac{-5 \pm \sqrt{25 + 400}}{-20} = \frac{-5 \pm \sqrt{425}}{-20}$

Simplify: $t = \frac{-5 \pm 20.62}{-20}$

We get two solutions: $t = \frac{-5 + 20.62}{-20} \quad \text{or} \quad t = \frac{-5 - 20.62}{-20}$

Only the positive root makes physical sense: $t = \frac{-5 + 20.62}{-20} = \frac{15.62}{-20} = -0.781$

Thus, the diver hits the water at approximately 0.78 seconds.

Part (c): Rearrange $h(t)$ into the form $A - B(t - C)^2$

We need to complete the square for the given quadratic function: $h(t) = 5t - 10t^2 + 10$ Rearrange terms: $h(t) = -10t^2 + 5t + 10$

To complete the square, factor out $-10$ from the first two terms: $h(t) = -10(t^2 - \frac{1}{2}t) + 10$

Now, complete the square inside the parentheses. Take half of $\frac{1}{2}$, square it, and add/subtract it: $h(t) = -10\left(t^2 - \frac{1}{2}t + \frac{1}{16} - \frac{1}{16}\right) + 10$ Simplify: $h(t) = -10\left(\left(t - \frac{1}{4}\right)^2 - \frac{1}{16}\right) + 10$

Distribute the $-10$: $h(t) = -10\left(t - \frac{1}{4}\right)^2 + \frac{10}{16} + 10$ Simplify further: $h(t) = -10\left(t - \frac{1}{4}\right)^2 + 10.625$

Thus, we have the form: $A = 10.625, \quad B = 10, \quad C = \frac{1}{4}$

Part (d): Using your answer to part (c), find the maximum height of the diver and the time at which this maximum height is reached.

From part (c), the maximum height occurs when the squared term is zero, i.e., at $t = C = \frac{1}{4}$.

The maximum height is the value of $h(t)$ at this time, which is $A = 10.625$ meters.

Answer:

• Maximum height = 10.625 meters
• Time at which maximum height is reached = 0.25 seconds

Would you like more details on any part of the solution? Here are some related questions for practice:

1. What is the physical interpretation of the constants $A$, $B$, and $C$?
2. How would the equation change if the springboard height was different?
3. Can you solve for the time to hit the water using numerical methods instead of the quadratic formula?
4. How can you derive the time of maximum height using calculus?
5. How does the negative coefficient of $t^2$ affect the shape of the trajectory?

Tip: Completing the square is useful for finding the vertex of a parabola, which corresponds to the maximum or minimum value of a quadratic function.