Calculus
posted by Abby .
If a ball is thrown straight up into the air with an initial velocity of 95 ft/s, its height in feet after t seconds is given by
f(t)=95t−16t^2
Find the average velocity for the time period beginning when t=1 and lasting
(i) 0.5 seconds
(ii) 0.1 seconds
(iii) 0.01 seconds
(iii) 0.001 seconds
Based on the above results, guess what the instantaneous velocity of the ball is when t=1.

in each case, the average velocity is ∆y/∆x for the given interval.
So, for [1,1.5],
∆y/∆x = (f(1.5)f(1))/(1.51)
= (106.579)/.5
= 55
Do the others similarly