The problem presented involves determining the position and velocity functions for a silver dollar dropped from a height of 1307 feet using the given position function for free-falling objects:

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

Where:

• $s(t)$ is the position of the object at time $t$,
• $v_0$ is the initial velocity,
• $s_0$ is the initial position, and
• $-16t^2$ represents the acceleration due to gravity (in feet per second squared).

Solution:

1. Position Function $s(t)$:

Since the silver dollar is dropped (and not thrown), the initial velocity $v_0 = 0$. The initial position $s_0 = 1307$ feet.

Thus, the position function simplifies to: $s(t) = -16t^2 + 1307$

This function correctly represents the height of the silver dollar at any time $t$.

2. Velocity Function $v(t)$:

The velocity function is the derivative of the position function $s(t)$.

$v(t) = \frac{d}{dt}[-16t^2 + 1307]$

Taking the derivative: $v(t) = -32t$

This is the velocity of the silver dollar at any time $t$.

Conclusion:

• The position function is $s(t) = -16t^2 + 1307$,
• The velocity function is $v(t) = -32t$.

The entries marked with red "X" in the image appear incorrect due to misunderstanding of initial conditions:

• The position function must include the correct height and form.
• The velocity at $t = 0$ is not zero, but depends on $v(t) = -32t$.

Would you like further clarification, or have any questions?

Here are 5 related questions to further expand on this topic:

1. What would the velocity of the silver dollar be after 2 seconds?
2. How long will it take for the silver dollar to reach the ground?
3. What is the maximum velocity of the silver dollar just before hitting the ground?
4. How would the position function change if the initial velocity was not zero?
5. Can you derive the time it takes for an object to fall from any height?

Tip: Always remember that velocity is the derivative of position, and acceleration is the derivative of velocity!