# Find the average velocity of the ball over the time interval from 4 to 4+h seconds when h cant = 0?  When a ball is thrown vertically upwardinto the air with a velocity of 67 ft/sec itsheight, y(t), in feet after t seconds is given byy(t) = 67t − 16t^2 .

By definition, the average velocity on the time interval `[t_1,t_2]` is

`(y(t_2)-y(t_1))/(t_2-t_1)` . In this case, we are given that `t_1=4` seconds and `t_2=4+h` seconds. We plug these into the given equation to get

`y(4)=67*4-16*4^2=268-256=12` feet and

`y(4+h)=67(4+h)-16(4+h)^2=268+67h-256-128h-16h^2`

`=-16h^2-61h+12`  feet. Now we just plug these into the average velocity formula:

`(y(t_2)-y(t_1))/(t_2-t_1)=((-16h^2-61h+12)-12)/((4+h)-4)`

`=(-16h^2-61h)/h=-16h-61`

The units in the numerator are feet and the units in the denominator are seconds, so the units of velocity turn out to be feet/second, as expected. The reason h can't be 0 is to avoid division by zero.

The average velocity on the interval [4,4+h] is -16h-61 ft/s.

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