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.

1 answer

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))/(1.5-1)
= (106.5-79)/.5
= 55

Do the others similarly