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Let's break down the problem using the given equation for the ball's height:

The height \( y(t) \) of the ball at time \( t \) is given by:

\[
y(t) = 29t - 14t^2
\]

### Part (a): Finding the average velocity over different time intervals
The average velocity over a time interval \([t_1, t_2]\) is computed as:

\[
\text{Average velocity} = \frac{y(t_2) - y(t_1)}{t_2 - t_1}
\]

Here, \( t_1 = 3 \), and \( t_2 = t_1 + \Delta t \), where \( \Delta t \) is the small time difference provided in the problem.

#### Let's compute each case for different values of \( \Delta t \):

1. For \( \Delta t = 0.01 \) s, \( t_2 = 3 + 0.01 = 3.01 \)
   \[
   \text{Average velocity} = \frac{y(3.01) - y(3)}{0.01}
   \]

2. For \( \Delta t = 0.005 \) s, \( t_2 = 3 + 0.005 = 3.005 \)
   \[
   \text{Average velocity} = \frac{y(3.005) - y(3)}{0.005}
   \]

3. For \( \Delta t = 0.002 \) s, \( t_2 = 3 + 0.002 = 3.002 \)
   \[
   \text{Average velocity} = \frac{y(3.002) - y(3)}{0.002}
   \]

4. For \( \Delta t = 0.001 \) s, \( t_2 = 3 + 0.001 = 3.001 \)
   \[
   \text{Average velocity} = \frac{y(3.001) - y(3)}{0.001}
   \]

I'll compute these values now.The average velocities for the different time intervals, rounded to three decimal places, are:

1. For \( \Delta t = 0.01 \) s: \( -55.140 \, \text{ft/s} \)
2. For \( \Delta t = 0.005 \) s: \( -55.070 \, \text{ft/s} \)
3. For \( \Delta t = 0.002 \) s: \( -55.028 \, \text{ft/s} \)
4. For \( \Delta t = 0.001 \) s: \( -55.014 \, \text{ft/s} \)

### Part (b): Estimating the instantaneous velocity at \( t = 3 \)

The instantaneous velocity is the derivative of the height function \( y(t) \) with respect to time \( t \). The velocity function is:

\[
v(t) = \frac{dy}{dt} = 29 - 28t
\]

Let's compute the velocity at \( t = 3 \):

\[
v(3) = 29 - 28 \times 3 = 29 - 84 = -55 \, \text{ft/s}
\]

So, the instantaneous velocity at \( t = 3 \) is \( -55 \, \text{ft/s} \).

Would you like further details or explanations on any part? Here are 5 related questions for practice:

1. How do you compute the average velocity over any arbitrary time interval?
2. Why does the instantaneous velocity at \( t = 3 \) match the limit of the average velocities?
3. What does a negative velocity indicate about the motion of the ball?
4. Can you compute the time when the ball reaches its maximum height using the velocity function?
5. What would the average velocity be if the time interval started at \( t = 4 \)?

**Tip**: The instantaneous velocity is the slope of the tangent to the curve at a given point, while average velocity is the slope of the secant line between two points.

A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 29 ft/s. Its height (in feet) is given by y = 29t - 14t^2, where t is measured in seconds. a. Find the average velocity (to three decimal places) for the time period beginning when t = 3 and lasting for: i. 0.01 s, ii. 0.005 s, iii. 0.002 s, iv. 0.001 s. b. Estimate the instantaneous velocity of the ball when t = 3.

Find the average and instantaneous velocity for a ball thrown into the air by a baby alien in Alpha Centauri. Learn how to compute velocity over various time intervals, as well as estimate instantaneous velocity using differentiation. Suitable for Grade 10-12 students.

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