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If a ball is thrown straight up into the air with an initial velocity of 95 ft/s, its height in feet after seconds is given

If a ball is thrown straight up into the air with an initial velocity of 95 ft/s, its height in feet after seconds is given by y= 95t - 16t^2. Find the average velocity (i.e. the change in distance with respect to the change in time) for the time period beginning when t=2 and lasting (i) 0.5 seconds: (ii) 0.1 seconds: (iii) 0.01 seconds: (iv) 0.0001 seconds: Finally, based on the above results, guess what the instantaneous velocity of the ball is when t=2.

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