If a ball is thrown straight up into the air with an initial velocity of 60 ft/s, its height in feet after 𝑡 seconds is given by 𝑦=60𝑡−16𝑡2. Find the average velocity (i.e. the change in distance with respect to the change in time) for the time period beginning when 𝑡=1 and lasting
(i) 0.5 seconds: 

(ii) 0.1 seconds: 

(iii) 0.01 seconds: 

(iv) 0.0001 seconds: 

We are given the equation for the height of the ball as a function of time:

\[
y(t) = 60t - 16t^2
\]

The average velocity over a time interval \([t_1, t_2]\) can be calculated using the formula:

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

For all parts of the problem, we are asked to find the average velocity starting at \(t_1 = 1\) second, but with different durations (\(t_2 = 1 + \Delta t\)).

### Part (i) 0.5 seconds:

\[
t_2 = 1 + 0.5 = 1.5
\]

Substitute \( t_1 = 1 \) and \( t_2 = 1.5 \) into the height equation:

\[
y(1) = 60(1) - 16(1)^2 = 60 - 16 = 44 \, \text{feet}
\]

\[
y(1.5) = 60(1.5) - 16(1.5)^2 = 90 - 16(2.25) = 90 - 36 = 54 \, \text{feet}
\]

The average velocity is:

\[
\text{Average Velocity} = \frac{y(1.5) - y(1)}{1.5 - 1} = \frac{54 - 44}{0.5} = \frac{10}{0.5} = 20 \, \text{ft/s}
\]

### Part (ii) 0.1 seconds:

\[
t_2 = 1 + 0.1 = 1.1
\]

Now, calculate the heights:

\[
y(1) = 44 \, \text{feet} \quad (\text{calculated earlier})
\]

\[
y(1.1) = 60(1.1) - 16(1.1)^2 = 66 - 16(1.21) = 66 - 19.36 = 46.64 \, \text{feet}
\]

The average velocity is:

\[
\text{Average Velocity} = \frac{y(1.1) - y(1)}{1.1 - 1} = \frac{46.64 - 44}{0.1} = \frac{2.64}{0.1} = 26.4 \, \text{ft/s}
\]

### Part (iii) 0.01 seconds:

\[
t_2 = 1 + 0.01 = 1.01
\]

Now, calculate the heights:

\[
y(1.01) = 60(1.01) - 16(1.01)^2 = 60.6 - 16(1.0201) = 60.6 - 16.3216 = 44.2784 \, \text{feet}
\]

The average velocity is:

\[
\text{Average Velocity} = \frac{y(1.01) - y(1)}{1.01 - 1} = \frac{44.2784 - 44}{0.01} = \frac{0.2784}{0.01} = 27.84 \, \text{ft/s}
\]

### Part (iv) 0.0001 seconds:

\[
t_2 = 1 + 0.0001 = 1.0001
\]

Now, calculate the heights:

\[
y(1.0001) = 60(1.0001) - 16(1.0001)^2 = 60.006 - 16(1.00020001) = 60.006 - 16.00320016 = 44.00279984 \, \text{feet}
\]

The average velocity is:

\[
\text{Average Velocity} = \frac{y(1.0001) - y(1)}{1.0001 - 1} = \frac{44.00279984 - 44}{0.0001} = \frac{0.00279984}{0.0001} = 27.9984 \, \text{ft/s}
\]

### Summary of Results:

- (i) Average velocity over 0.5 seconds: \( 20 \, \text{ft/s} \)
- (ii) Average velocity over 0.1 seconds: \( 26.4 \, \text{ft/s} \)
- (iii) Average velocity over 0.01 seconds: \( 27.84 \, \text{ft/s} \)
- (iv) Average velocity over 0.0001 seconds: \( 27.9984 \, \text{ft/s} \)

Would you like more details or have any questions?

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Here are 5 related questions:
1. What is the instantaneous velocity of the ball at \(t = 1\) second?
2. How does the average velocity compare to the instantaneous velocity as the time interval shrinks?
3. What is the maximum height reached by the ball?
4. How long does the ball take to reach the maximum height?
5. What is the total time the ball stays in the air before hitting the ground?

**Tip:** As the time interval decreases, the average velocity approaches the instantaneous velocity at that point.

If a ball is thrown straight up into the air with an initial velocity of 60 ft/s, its height in feet after t seconds is given by y=60t−16t^2. Find the average velocity (i.e., the change in distance with respect to the change in time) for the time period beginning when t=1 and lasting (i) 0.5 seconds, (ii) 0.1 seconds, (iii) 0.01 seconds, and (iv) 0.0001 seconds.

Solve for the average velocity of a ball thrown straight up with an initial velocity of 60 ft/s using the equation y=60t−16t^2. This problem involves calculating the average velocity over different time intervals, applying kinematics and quadratic equations. Suitable for high school students, Grades 10-12.

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