Question

find average velocity between 2 to 4 sec

Answer 1

First you find the height after 2 seconds, we'll denote that as $y_2$. So using the fomula we have:

$$y_2 = 40 \cdot 2 - 16 \cdot 2^2 = 80 - 64 = 16$$

After 0.5 seconds, the time passed after the throw will be 2.5 seconds so for this height we have:

$y_{2.5} = 40\frac 52 - 16 \left(\frac 52\right)^2 =100 - 100 = 0$$

So after 2.5 second the height of the ball will be $0\text{ ft}$.

We know that the average velocity can be calulated using the following formula:

$$v = \frac{\Delta s}{\Delta t} = \frac{|y_1 - y_0|}{|t_1 - t_0|}$$

So after the substitution we have:

$$v = \frac{|0 - 16|}{|2.5 - 2|} = \frac{|-16|}{|0.5|} = \frac{16}{0.5} = 32$$

So the average speed in that period will be 32 $\frac{ft}{s}$

Now you can do the 0.1 second interval by yourself.