Question

If a ball is thrown straight up into the air with an initial velocity of 40 ft/s, it height in feet after t second is given by y=40t−16t2. Find the average velocity for the time period begining when t=2 and lasting
(i) 0.5 seconds

(ii) 0.1 seconds

(iii) 0.01 seconds

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

Answer 1

We should first calculate the highest point that ball reaches. 
y' = 40 - 32t = 0
t = 40/32 = 1,25s
y = 25 feet.

To calculate average velocity we use simple formula:
Vav=s/t   where s is traveled distance by the time t.



for t=2  we calculate y
y = 16
(i)  for t = 2,5     y = 0
Vav = 16/0.5 = 32 feet/s
(ii) for t = 2.1     y = 13.44
Vav = 2.56/0.1 = 25.6 feet/s
(iii) for t = 2.01  y = 15.7584
Vav = 0.2416/0.01= 24.16 feet/s

it seems like the answer would be 24 feet/s. There is a way to calculate that.