A graphing calculator is recommended.
If a ball is thrown into the air with an initial velocity of 42 ft/s, its height in feet after t seconds is given by 
y = 42t − 16t2.
(a)
Find the average velocity of the ball (in ft/s) for the time interval beginning at t = 2 and lasting for each of the following.
(i)
0.5 seconds
-30
 
Correct: Your answer is correct.
 ft/s
(ii)
0.1 seconds
-23.6
 
Correct: Your answer is correct.
 ft/s
(iii)
0.05 seconds
18.26
 
Changed: Your submitted answer was incorrect. Your current answer has not been submitted.
 ft/s
(iv)
0.01 seconds
-34.8
 
Changed: Your submitted answer was incorrect. Your current answer has not been submitted.
 ft/s
(b)
Use your answers from part (a) to estimate the instantaneous velocity (in ft/s) when 
t = 2.
 ft/s

Let's break down the problem step by step to find the average velocities for the specified time intervals and estimate the instantaneous velocity at \( t = 2 \).

### The given equation for height as a function of time is:
\[
y = 42t - 16t^2
\]
This equation represents the height of the ball in feet after \( t \) seconds, where \( t \geq 0 \).

### (a) Average velocity over different time intervals

The formula for **average velocity** over a time interval from \( t \) to \( t + \Delta t \) is:
\[
v_{\text{avg}} = \frac{y(t + \Delta t) - y(t)}{\Delta t}
\]
where \( \Delta t \) is the duration of the time interval.

The height function is \( y(t) = 42t - 16t^2 \), so we will substitute this into the formula for average velocity for each of the given intervals, starting at \( t = 2 \).

#### (i) Time interval: 0.5 seconds (\( \Delta t = 0.5 \))
We need to compute:
\[
v_{\text{avg}} = \frac{y(2 + 0.5) - y(2)}{0.5}
\]

1. \( y(2.5) = 42(2.5) - 16(2.5)^2 = 105 - 16(6.25) = 105 - 100 = 5 \)
2. \( y(2) = 42(2) - 16(2)^2 = 84 - 16(4) = 84 - 64 = 20 \)

Now, compute the average velocity:
\[
v_{\text{avg}} = \frac{5 - 20}{0.5} = \frac{-15}{0.5} = -30 \, \text{ft/s}
\]
The answer for this interval is \( -30 \, \text{ft/s} \).

#### (ii) Time interval: 0.1 seconds (\( \Delta t = 0.1 \))
We need to compute:
\[
v_{\text{avg}} = \frac{y(2 + 0.1) - y(2)}{0.1}
\]

1. \( y(2.1) = 42(2.1) - 16(2.1)^2 = 88.2 - 16(4.41) = 88.2 - 70.56 = 17.64 \)
2. \( y(2) = 20 \) (as calculated above)

Now, compute the average velocity:
\[
v_{\text{avg}} = \frac{17.64 - 20}{0.1} = \frac{-2.36}{0.1} = -23.6 \, \text{ft/s}
\]
The answer for this interval is \( -23.6 \, \text{ft/s} \).

#### (iii) Time interval: 0.05 seconds (\( \Delta t = 0.05 \))
We need to compute:
\[
v_{\text{avg}} = \frac{y(2 + 0.05) - y(2)}{0.05}
\]

1. \( y(2.05) = 42(2.05) - 16(2.05)^2 = 86.1 - 16(4.2025) = 86.1 - 67.24 = 18.86 \)
2. \( y(2) = 20 \) (as calculated above)

Now, compute the average velocity:
\[
v_{\text{avg}} = \frac{18.86 - 20}{0.05} = \frac{-1.14}{0.05} = -22.8 \, \text{ft/s}
\]
So the answer for this interval should be \( -22.8 \, \text{ft/s} \).

#### (iv) Time interval: 0.01 seconds (\( \Delta t = 0.01 \))
We need to compute:
\[
v_{\text{avg}} = \frac{y(2 + 0.01) - y(2)}{0.01}
\]

1. \( y(2.01) = 42(2.01) - 16(2.01)^2 = 84.42 - 16(4.0401) = 84.42 - 64.64 = 19.78 \)
2. \( y(2) = 20 \) (as calculated above)

Now, compute the average velocity:
\[
v_{\text{avg}} = \frac{19.78 - 20}{0.01} = \frac{-0.22}{0.01} = -22 \, \text{ft/s}
\]
So the answer for this interval should be \( -22 \, \text{ft/s} \).

### (b) Instantaneous velocity at \( t = 2 \)

The instantaneous velocity is the derivative of the height function \( y(t) \) with respect to \( t \):
\[
v(t) = \frac{dy}{dt} = 42 - 32t
\]

At \( t = 2 \):
\[
v(2) = 42 - 32(2) = 42 - 64 = -22 \, \text{ft/s}
\]
The instantaneous velocity when \( t = 2 \) is \( -22 \, \text{ft/s} \).

---

Would you like further details on how these calculations were done or explanations of specific parts of the process?

### Related Questions:
1. What does the negative sign in velocity indicate in this problem?
2. How does the formula for average velocity differ from instantaneous velocity?
3. Can you explain the concept of a derivative in relation to instantaneous velocity?
4. What is the significance of the height function being quadratic?
5. How does changing the initial velocity affect the height function?

**Tip:** When estimating instantaneous velocity, smaller time intervals yield more accurate results!

If a ball is thrown into the air with an initial velocity of 42 ft/s, its height in feet after t seconds is given by y = 42t − 16t^2. (a) Find the average velocity of the ball (in ft/s) for the time interval beginning at t = 2 and lasting for each of the following: (i) 0.5 seconds, (ii) 0.1 seconds, (iii) 0.05 seconds, (iv) 0.01 seconds. (b) Use your answers from part (a) to estimate the instantaneous velocity (in ft/s) when t = 2.

Explore how to calculate the average and instantaneous velocity of a ball thrown with an initial velocity of 42 ft/s using the quadratic height function y = 42t - 16t^2. Step-by-step solutions for multiple time intervals and an explanation of derivative-based velocity calculations. Suitable for Grades 10-12.

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