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The problem in the image involves a ball thrown straight up with an initial velocity of \( 43 \, \text{ft/s} \). The height of the ball after \( t \) seconds is described by the equation:

\[
y = 43t - 16t^2
\]

You are asked to find the average velocity for the time period from \( t = 1 \) to \( t = 1 + h \) for various values of \( h \), where \( h \) is given as \( 0.01 \, \text{sec}, 0.005 \, \text{sec}, 0.002 \, \text{sec}, 0.001 \, \text{sec} \).

### Step-by-step approach:

1. **Average velocity formula:**
   The average velocity \( v_{\text{avg}} \) over a time interval from \( t_1 \) to \( t_2 \) is given by:
   \[
   v_{\text{avg}} = \frac{y(t_2) - y(t_1)}{t_2 - t_1}
   \]
   In this case, \( t_1 = 1 \) and \( t_2 = 1 + h \). So the formula becomes:
   \[
   v_{\text{avg}} = \frac{y(1 + h) - y(1)}{h}
   \]
   
2. **Calculate \( y(1) \):**
   Using the given equation for \( y(t) \):
   \[
   y(1) = 43(1) - 16(1)^2 = 43 - 16 = 27 \, \text{ft}
   \]

3. **Calculate \( y(1 + h) \):**
   We substitute \( t = 1 + h \) into the equation for \( y(t) \):
   \[
   y(1 + h) = 43(1 + h) - 16(1 + h)^2
   \]
   Expanding the terms:
   \[
   y(1 + h) = 43(1 + h) - 16(1 + 2h + h^2) = 43 + 43h - 16(1 + 2h + h^2)
   \]
   \[
   y(1 + h) = 43 + 43h - 16(1 + 2h + h^2) = 43 + 43h - 16 - 32h - 16h^2
   \]
   Simplifying:
   \[
   y(1 + h) = 27 + (43 - 32)h - 16h^2 = 27 + 11h - 16h^2
   \]

4. **Substitute into the average velocity formula:**
   Now that we have expressions for \( y(1 + h) \) and \( y(1) \), the average velocity becomes:
   \[
   v_{\text{avg}} = \frac{(27 + 11h - 16h^2) - 27}{h} = \frac{11h - 16h^2}{h} = 11 - 16h
   \]

5. **Compute the average velocity for each value of \( h \):**

   - For \( h = 0.01 \):
     \[
     v_{\text{avg}} = 11 - 16(0.01) = 11 - 0.16 = 10.84 \, \text{ft/s}
     \]
   
   - For \( h = 0.005 \):
     \[
     v_{\text{avg}} = 11 - 16(0.005) = 11 - 0.08 = 10.92 \, \text{ft/s}
     \]
   
   - For \( h = 0.002 \):
     \[
     v_{\text{avg}} = 11 - 16(0.002) = 11 - 0.032 = 10.968 \, \text{ft/s}
     \]
   
   - For \( h = 0.001 \):
     \[
     v_{\text{avg}} = 11 - 16(0.001) = 11 - 0.016 = 10.984 \, \text{ft/s}
     \]

### Final answers:
- \( h = 0.01 \, \text{sec} \): \( v_{\text{avg}} = 10.84 \, \text{ft/s} \)
- \( h = 0.005 \, \text{sec} \): \( v_{\text{avg}} = 10.92 \, \text{ft/s} \)
- \( h = 0.002 \, \text{sec} \): \( v_{\text{avg}} = 10.968 \, \text{ft/s} \)
- \( h = 0.001 \, \text{sec} \): \( v_{\text{avg}} = 10.984 \, \text{ft/s} \)

Would you like any further details on this? Here are 5 related questions to consider:

1. How does the average velocity formula relate to the concept of instantaneous velocity?
2. How would the velocity change if the initial velocity were different?
3. Can we derive the instantaneous velocity by taking the limit as \( h \to 0 \)?
4. How does the quadratic term in the height equation influence the motion of the ball?
5. What would the average velocity be for larger intervals of \( h \), such as \( h = 0.1 \)?

**Tip:** As \( h \) decreases, the average velocity approaches the instantaneous velocity at \( t = 1 \).

A ball is thrown straight up into the air with an initial velocity of 43 ft/s. Its height in feet after t seconds is given by y = 43t - 16t^2. Find the average velocity for the time period from t = 1 to t = 1 + h, when h is given as: 0.01 sec, 0.005 sec, 0.002 sec, and 0.001 sec.

Calculate the average velocity of a ball thrown straight up with an initial velocity of 43 ft/s. This problem explores kinematic equations in vertical motion, using the formula for height y = 43t - 16t^2, and covers the concept of average velocity for different small time intervals.

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