# calculus

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 begining when t = 1 and lasting
(i) 0.1 seconds

(ii) 0.01 seconds

(iii) 0.001 seconds

Finally based on the above results, guess what the instantaneous velocity of the ball is when t =1

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1. So, the question is stating that the time period you are looking for is when the ball is thrown at 1 second and goes to .1 second. So, you want the velocity of the ball at the interval [1,1.1]. Velocity is the change in distance/ change in time. What is the change of the distance of the change in time from 1 second to 1.1 second? If the equation of f(x)= 70 t-16 t^2, the f(x) is the distance of the ball, so you should see that f(1.1)-f(1) is the change in height. Everything together is the average rate of change formula, where [f(1.1)-f(1)]/1.1-1=the average velocity of the ball. Now, you can use the formula to make the interval even smaller.

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