The **average velocity** for the time period beginning when t=1 and lasting

(i) 0.01 seconds = **63.84 ft/s**

(ii) 0.001 seconds =** 63.984 ft/s**

Given that a ball is thrown with an** initial velocity **= 80 ft/s

Let** 'y' **be the height in feet after** 't'** seconds.

Given, gives the height in 't' seconds.

**Average velocity** = Rate of change of distance

= Change in distance/Change in time.

The initial time can be taken as 0 s.

When** t =1 s**, y = 80 - 16 = 64 ft

(1) **t = 0.01 s**

y = 80 x 0.01 - 16 x 0.01 x 0.01 = 0.7984 ft

** Average velocity** = (64 - 0.7984) / (1 -0.01) =** 63.84 ft/s**

(2) **t = 0.001 s**

y = 80 x 0.001 - 16 x 0.001 x 0.001 = 0.079984 ft

** Average velocity** = (64 - 0.079984) / (1 -0.001) =** 63.984 ft/s**

The question is incomplete. Find out the complete question below:

If a ball is thrown straight up into the air with an initial velocity of 80 ft/s, it height in feet after t second is given by .Find the average velocity for the time period beginning when t=1 and lasting

(i) 0.01 seconds

(ii) 0.001 seconds

Learn more about **average velocity **at brainly.com/question/6504879

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