If a ball is thrown straight up into the air with an initial velocity of 40 ft/s, its height in feet after seconds is given by y=40−162. Find the average velocity (i.e. the change in distance with respect to the change in time) for the time period beginning when =2 and lasting
(i) 0.5 seconds:
(ii) 0.1 seconds:
(iii) 0.01 seconds:
(iv) 0.0001 seconds:
Finally, based on the above results, guess what the instantaneous velocity of the ball is when =2.
Answer:
