If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, it height in feet after t second is given by y = 70 t - 16 t^2. Find the average velocity for the time period beginning when t = 2 and lasting

(i) 0.1 seconds

Velocity = displacement / time = y2-y1/t2-t1

y = 70t-16t^2 we know t = 2 and initial velocity = 70

so av = y2 (value of y when we sub in t=2) which for this case y=70 (2) - 16 (2^2) = 76

therefor 76-70/2.1-2 = 60 a)

and 76-70/2.01-2 = 600 b)

and 76-70/2.001-2=6000 c)

a) 76-70/0.1 = 60ft

b) 76-70/0.01 = 600ft

c) 76-70/0.001 = 6000ft