Question

If a ball is thrown straight up into the air with an initial velocity of 50 ft/s, it height in feet after t second is given by y = -50 t - 16 t^2. Find the average velocity for the time period begining when t = 2 and lasting
(i) 0.1 seconds

(ii) 0.01 seconds

(iii) 0.001 seconds

Finally based on the above results, guess what the instantaneous velocity of the ball is when t =2

Answer 1

To find the average velocity of the ball over the specified time periods, we can use the formula for average velocity over an interval \([t, t + h]\):

\[
\text{Average Velocity} = \frac{y(t + h) - y(t)}{h}
\]

Given the height function:

\[
y(t) = -16t^2 + 50t
\]

We will compute \(y(2)\) first:

\[
y(2) = -16(2)^2 + 50(2) = -16(4) + 100 = -64 + 100 = 36 \text{ feet}
\]

**Now we will calculate average velocity for each case:**

### (i) For \(h = 0.1\) seconds (i.e., from \(t = 2\) to \(t = 2.1\))

Calculate \(y(2.1)\):

\[
y(2.1) = -16(2.1)^2 + 50(2.1) = -16(4.41) + 105 = -70.56 + 105 = 34.44 \text{ feet}
\]

Now, compute the average velocity:

\[
\text{Average Velocity} = \frac{y(2.1) - y(2)}{0.1} = \frac{34.44 - 36}{0.1} = \frac{-1.56}{0.1} = -15.6 \text{ ft/s}
\]

### (ii) For \(h = 0.01\) seconds (i.e., from \(t = 2\) to \(t = 2.01\))

Calculate \(y(2.01)\):

\[
y(2.01) = -16(2.01)^2 + 50(2.01) = -16(4.0401) + 100.5 = -64.6416 + 100.5 = 35.8584 \text{ feet}
\]

Calculate the average velocity:

\[
\text{Average Velocity} = \frac{y(2.01) - y(2)}{0.01} = \frac{35.8584 - 36}{0.01} = \frac{-0.1416}{0.01} = -14.16 \text{ ft/s}
\]

### (iii) For \(h = 0.001\) seconds (i.e., from \(t = 2\) to \(t = 2.001\))

Calculate \(y(2.001)\):

\[
y(2.001) = -16(2.001)^2 + 50(2.001) = -16(4.004001) + 100.05 = -64.064016 + 100.05 = 35.985984 \text{ feet}
\]

Calculate the average velocity:

\[
\text{Average Velocity} = \frac{y(2.001) - y(2)}{0.001} = \frac{35.985984 - 36}{0.001} = \frac{-0.014016}{0.001} = -14.016 \text{ ft/s}
\]

### Summary of Average Velocities
- Average Velocity for \(h = 0.1\): \(-15.6 \text{ ft/s}\)
- Average Velocity for \(h = 0.01\): \(-14.16 \text{ ft/s}\)
- Average Velocity for \(h = 0.001\): \(-14.016 \text{ ft/s}\)

### Instantaneous Velocity at \(t = 2\)

Based on the results from the average velocities as \(h\) approaches 0, it appears the instantaneous velocity at \(t = 2\) is trending toward \(-14 \text{ ft/s}\).

To verify this analytically, we can find the instantaneous velocity by taking the derivative of \(y(t)\):

\[
\frac{dy}{dt} = -32t + 50
\]

At \(t = 2\):

\[
\frac{dy}{dt}\Big|_{t=2} = -32(2) + 50 = -64 + 50 = -14 \text{ ft/s}
\]

Thus, the instantaneous velocity at \(t = 2\) is:

\[
\boxed{-14 \text{ ft/s}}
\]