If a ball is thrown into the area (basic calculus...please no mention of derivatives)

I need to know how to set up the formula for this:

If a ball is thrown into the air with a velocity of 40 ft/s, its height in feet is t seconds later is given by y=40t - 16t^2.

a) Find the average velocity for the time period begining when t=2 lasting
i) .5 second
ii) .1 second
iii) .05 second
iv) .01 second


what do I do?

These are all done the same way, so I will do just part i.

y = 40t - 16t^2

[y(2.5) - y(2)]/(2.5 - 2) = {[40(2.5) - 16(2.5)^2] - [40(2) - 16(2)^2]}/0.5 = (0 - 16)/0.5 = -16/0.5 = -32

Average velocity is the sum of velocities over the time of interest divided by the time. So:
integral(v dt) / time

But the integral of v over time is just the y(t) you have above. So
average velocity = [y(time2) - y(time1)] / (time2-time1)

For example, the velocity starting with t=0 for 0.1 second is:
[40*0.1 - 16*0.1^2 - (40*0 - 16*0^2)] / 0.1 = 40 - 16*0.1 = 40 - 1.6 = 38.4

This shows the velocity right after throwing it up is close to the 40 ft/s as expected. Using other values for time1 and time2 can provide you the answers you need.