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 $-22 \, \text{ft/s}$.

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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!