If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t second is given by y=70t−16t2. Find the average velocity for the time period beginning when t = 2 and lasting:
(i) 0.1 seconds
(ii) 0.01 seconds and
(iii) 0.001 seconds.

i.4.4m/s

ii.5.8m/s

III.5.98m/s

Explanation:

Given Y=70t-16t^2 at t=2

Y= 70(2)-16(2)^2=76m/s

i.at 0.1sec=2.1secs

Y=70(2.1)-16(2.1)^2=76.44m/s

Velocity (76.44-76/.1)=4.4m/s

ii.at t=0.01secs

Y=70(2.01)-16(2.01)^2=76.058m/s

Velocity=(76.056-76/.01)=5.8m/s

ii.at t=0.001

Y=70(2.001)-16(2.001)^2=76.0059m/s

Velocity=(76.0059-76/.001)=5.98m/s

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