Let's work through each part of the problem step by step.

Part (a): Position and velocity functions

The given position function for a free-falling object is:

$s(t) = -16t^2 + v_0t + s_0$

where:

• $s(t)$ is the position at time $t$ (in feet),
• $v_0$ is the initial velocity (in ft/s),
• $s_0$ is the initial height (in feet),
• and $t$ is the time (in seconds).

For this problem:

• The coin is dropped from a height of 1,310 feet, so the initial velocity $v_0 = 0$ ft/s (because it's dropped, not thrown).
• The initial height $s_0 = 1310$ feet.

Thus, the position function becomes:

$s(t) = -16t^2 + 1310$

To find the velocity function, we differentiate the position function with respect to $t$:

$v(t) = \frac{d}{dt} \left( -16t^2 + 1310 \right) = -32t$

So:

• The position function is $s(t) = -16t^2 + 1310$,
• The velocity function is $v(t) = -32t$.

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Part (b): Average velocity on the interval [3, 4]

The average velocity $\text{v}_{\text{avg}}$ on the interval $[a, b]$ is given by:

$\text{v}_{\text{avg}} = \frac{s(b) - s(a)}{b - a}$

For the interval $[3, 4]$:

1. Find $s(3)$ and $s(4)$: $s(3) = -16(3)^2 + 1310 = -144 + 1310 = 1166 \, \text{feet}$ $s(4) = -16(4)^2 + 1310 = -256 + 1310 = 1054 \, \text{feet}$

2. Calculate the average velocity: $\text{v}_{\text{avg}} = \frac{1054 - 1166}{4 - 3} = \frac{-112}{1} = -112 \, \text{ft/s}$

This is correct as provided in the problem.

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Part (c): Instantaneous velocity at $t = 3$ and $t = 4$

The instantaneous velocity is given by the velocity function $v(t) = -32t$.

For $t = 3$: $v(3) = -32(3) = -96 \, \text{ft/s}$

For $t = 4$: $v(4) = -32(4) = -128 \, \text{ft/s}$

These are also correct as provided.

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Part (d): Time to reach the ground

To find the time when the coin reaches the ground, set $s(t) = 0$ (since the ground level is at height 0):

$-16t^2 + 1310 = 0$

Solve for $t$:

$16t^2 = 1310$ $t^2 = \frac{1310}{16} = 81.875$ $t = \sqrt{81.875} \approx 9.049 \, \text{seconds}$

Thus, the time required for the coin to reach the ground is approximately $t \approx 9.049$ seconds.

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Part (e): Velocity at impact

To find the velocity at impact, use the velocity function $v(t) = -32t$, and substitute $t = 9.049$:

$v(9.049) = -32(9.049) \approx -289.6 \, \text{ft/s}$

Thus, the velocity at impact is approximately $-289.6 \, \text{ft/s}$.

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Summary of Results

• Position function: $s(t) = -16t^2 + 1310$
• Velocity function: $v(t) = -32t$
• Average velocity on [3, 4]: $-112 \, \text{ft/s}$
• Instantaneous velocities:
  • $v(3) = -96 \, \text{ft/s}$
  • $v(4) = -128 \, \text{ft/s}$
• Time to reach the ground: $t \approx 9.049 \, \text{seconds}$
• Velocity at impact: $v \approx -289.6 \, \text{ft/s}$

Would you like further clarification or details on any part?

Here are 5 related questions you could explore:

1. What would happen if the initial velocity wasn't zero?
2. How would air resistance affect the results?
3. What is the physical meaning of negative velocity in this problem?
4. Can we determine the maximum height if the coin were thrown upwards instead of dropped?
5. How does the time to fall change if the height is doubled?

Tip: When solving for time in quadratic equations, always consider both positive and negative roots, but remember that negative time is not physically meaningful in this context!