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### 14. A quarter is dropped from a height of 144 feet. Its height \( s(t) \) in feet, \( t \) seconds after being dropped, is given by:
\[
s(t) = 144 - 16t^2
\]

#### (a) Show algebraically that the quarter will hit the ground 3 seconds after it has been dropped.

The quarter hits the ground when the height \( s(t) = 0 \). We solve for \( t \) when \( s(t) = 0 \):

\[
s(t) = 144 - 16t^2 = 0
\]
\[
144 = 16t^2
\]
\[
t^2 = \frac{144}{16} = 9
\]
\[
t = \sqrt{9} = 3 \text{ seconds}
\]

So, the quarter hits the ground at \( t = 3 \) seconds.

---

#### (b) Find the average velocity of the quarter on each of the given time intervals.

The formula for average velocity between two time intervals \( t_1 \) and \( t_2 \) is:
\[
\text{Average velocity} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}
\]

##### (i) Time interval [2.5 sec, 3 sec]

For \( t_1 = 2.5 \) and \( t_2 = 3 \):
\[
s(2.5) = 144 - 16(2.5)^2 = 144 - 16(6.25) = 144 - 100 = 44
\]
\[
s(3) = 144 - 16(3)^2 = 144 - 144 = 0
\]
Now, the average velocity:
\[
\text{Average velocity} = \frac{0 - 44}{3 - 2.5} = \frac{-44}{0.5} = -88 \text{ feet per second}
\]

##### (ii) Time interval [2.9 sec, 3 sec]

For \( t_1 = 2.9 \) and \( t_2 = 3 \):
\[
s(2.9) = 144 - 16(2.9)^2 = 144 - 16(8.41) = 144 - 134.56 = 9.44
\]
\[
s(3) = 0
\]
Now, the average velocity:
\[
\text{Average velocity} = \frac{0 - 9.44}{3 - 2.9} = \frac{-9.44}{0.1} = -94.4 \text{ feet per second}
\]

##### (iii) Time interval [2.99 sec, 3 sec]

For \( t_1 = 2.99 \) and \( t_2 = 3 \):
\[
s(2.99) = 144 - 16(2.99)^2 = 144 - 16(8.9401) = 144 - 143.0416 = 0.9584
\]
\[
s(3) = 0
\]
Now, the average velocity:
\[
\text{Average velocity} = \frac{0 - 0.9584}{3 - 2.99} = \frac{-0.9584}{0.01} = -95.84 \text{ feet per second}
\]

---

#### (c) Estimate the instantaneous velocity of the quarter as it hits the ground using part (b).

From part (b), as the time intervals get smaller, the average velocity approaches -96 feet per second. Therefore, we estimate the instantaneous velocity to be:

\[
v(3) \approx -96 \text{ feet per second}
\]

---

#### (d) Use the limit definition of the derivative to find \( s'(3) \), the instantaneous velocity.

The instantaneous velocity is the derivative of \( s(t) \), i.e., \( v(t) = s'(t) \). We differentiate \( s(t) = 144 - 16t^2 \):

\[
s'(t) = \frac{d}{dt}[144 - 16t^2] = -32t
\]

At \( t = 3 \):
\[
s'(3) = -32(3) = -96 \text{ feet per second}
\]

Thus, the instantaneous velocity at \( t = 3 \) is \( -96 \) feet per second.

---

Let me know if you'd like further clarification on any part, and here are 5 follow-up questions to test your understanding:

1. What is the difference between average velocity and instantaneous velocity?
2. How would the problem change if the quarter was thrown upward instead of dropped?
3. What is the significance of the negative sign in the velocity values?
4. How can you calculate velocity at any other point during the fall?
5. Can you derive a formula for the time it takes an object to fall from a given height using only gravity?

### Tip:
When differentiating quadratic functions, always keep in mind that the derivative of \( at^2 \) is \( 2at \). This can help you speed up calculations when working with similar problems!

A quarter is dropped from a height of 144 feet. Its height, in feet, t seconds after being dropped is given by s(t) = 144 - 16t^2. Part (a) shows algebraically that the quarter will hit the ground 3 seconds after being dropped. Part (b) asks for the average velocity of the quarter over specific time intervals. Part (c) asks for an estimate of the instantaneous velocity as it hits the ground using answers from part (b). Part (d) involves using the limit definition of the derivative to calculate the instantaneous velocity.

This problem explores the motion of a quarter dropped from a height of 144 feet using quadratic functions. Learn to calculate the average and instantaneous velocity of the quarter as it falls, applying limits and derivatives. This problem is suitable for Grades 11-12 or early college students studying physics or calculus.

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