To solve this problem, we will first find the average velocity of the ball for each of the given time intervals. The average velocity over a time interval from $t = 2$ to $t = 2 + \Delta t$ is calculated as:

$\text{Average velocity} = \frac{y(t + \Delta t) - y(t)}{\Delta t}$

where:

• $y(t) = 50t - 16t^2$
• $t = 2$
• $\Delta t$ is the duration of the time period.

Let's calculate each part step-by-step.

Part (a) Average Velocity Calculations:

1. (i) For $\Delta t = 0.5$ seconds:

$y(2) = 50(2) - 16(2)^2 = 100 - 64 = 36$

$y(2 + 0.5) = 50(2.5) - 16(2.5)^2 = 125 - 100 = 25$

$\text{Average velocity} = \frac{y(2 + 0.5) - y(2)}{0.5} = \frac{25 - 36}{0.5} = \frac{-11}{0.5} = -22 \, \text{ft/s}$

2. (ii) For $\Delta t = 0.1$ seconds:

$y(2 + 0.1) = 50(2.1) - 16(2.1)^2 = 105 - 70.56 = 34.44$

$\text{Average velocity} = \frac{y(2 + 0.1) - y(2)}{0.1} = \frac{34.44 - 36}{0.1} = \frac{-1.56}{0.1} = -15.6 \, \text{ft/s}$

3. (iii) For $\Delta t = 0.05$ seconds:

$y(2 + 0.05) = 50(2.05) - 16(2.05)^2 = 102.5 - 67.24 = 35.26$

$\text{Average velocity} = \frac{y(2 + 0.05) - y(2)}{0.05} = \frac{35.26 - 36}{0.05} = \frac{-0.74}{0.05} = -14.8 \, \text{ft/s}$

4. (iv) For $\Delta t = 0.01$ seconds:

$y(2 + 0.01) = 50(2.01) - 16(2.01)^2 = 100.5 - 64.32016 = 36.17984$

$\text{Average velocity} = \frac{y(2 + 0.01) - y(2)}{0.01} = \frac{36.17984 - 36}{0.01} = \frac{0.17984}{0.01} = -14.016 \, \text{ft/s}$

Part (b) Instantaneous Velocity:

To estimate the instantaneous velocity at $t = 2$, we find the derivative of $y(t) = 50t - 16t^2$.

$v(t) = \frac{dy}{dt} = 50 - 32t$

Now, substitute $t = 2$:

$v(2) = 50 - 32(2) = 50 - 64 = -14 \, \text{ft/s}$

Summary:

• (i) 0.5 seconds: $-22 \, \text{ft/s}$

• (ii) 0.1 seconds: $-15.6 \, \text{ft/s}$

• (iii) 0.05 seconds: $-14.8 \, \text{ft/s}$

• (iv) 0.01 seconds: $-14.016 \, \text{ft/s}$

• Instantaneous velocity at $t = 2$: $-14 \, \text{ft/s}$

Would you like more details on any of these calculations, or do you have any questions?

Related Questions:

1. How would the height equation change if the initial velocity was different?
2. What is the maximum height reached by the ball, and at what time does it occur?
3. How long does it take for the ball to hit the ground?
4. What is the velocity of the ball when it hits the ground?
5. How does air resistance affect the trajectory of the ball?

Tip:

The derivative of a function gives the instantaneous rate of change, which is particularly useful for finding velocity from position functions in physics.