Step-by-step solution:

1. Equation to find the time when the diver hits the water:

The diver falls from a height of 49 ft, so at the moment of impact, the distance $d(t)$ will be equal to 49 ft.

Given the distance function: $d(t) = 16t^2$

To find when the diver hits the water, we set $d(t) = 49$ and solve for $t$: $16t^2 = 49$

Thus, the correct equation to solve for $t$ is: $16t^2 = 49$

2. Solving for $t$:

Now, solve the equation $16t^2 = 49$: $t^2 = \frac{49}{16}$ $t = \sqrt{\frac{49}{16}} = \frac{7}{4} = 1.75 \, \text{seconds}$

So, the diver will hit the water after 1.75 seconds.

3. Value of $a$ for calculating the velocity:

The formula for the velocity at time $t = a$ is given by the limit: $\lim_{h \to 0} \frac{d(a + h) - d(a)}{h}$

To find the velocity at the moment the diver hits the water, we use $t = a$ when the diver reaches the water. From the previous calculation, we know that the diver hits the water at $t = 1.75$ seconds. Therefore: $a = 1.75 \, \text{seconds}$

4. Value of $d(a)$ when the diver hits the water:

The value of $d(a)$ when the diver hits the water is the distance traveled at $t = 1.75$ seconds, which is the height from which the diver fell. So: $d(a) = 49 \, \text{ft}$

5. Velocity of the diver when they hit the water:

The velocity function is the derivative of the distance function $d(t) = 16t^2$, which gives: $v(t) = \frac{d}{dt}(16t^2) = 32t$

At $t = 1.75$ seconds, the velocity is: $v(1.75) = 32 \times 1.75 = 56 \, \text{ft/s}$

Thus, the velocity of the diver when they hit the water is 56 ft/s.

Final Answers:

• The equation that can be solved for $t$ is: $16t^2 = 49$
• The time when the diver hits the water is: 1.75 seconds
• The value of $a$ for calculating velocity is: 1.75 seconds
• The value of $d(a)$ when the diver hits the water is: 49 ft
• The velocity of the diver when they hit the water is: 56 ft/s

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Would you like further clarification on any of the steps, or have any other questions? Here's a list of related questions:

1. What would happen if the height from which the diver falls was different?
2. How does the velocity change with time for the diver?
3. What is the relationship between distance and velocity for free fall?
4. How can we use calculus to model other physical phenomena like this?
5. How would the diver’s velocity change if air resistance was included?

Tip: When solving quadratic equations, always check for both positive and negative solutions, but for time problems like this, only the positive solution makes physical sense.