Question

(1 point)
a ball is thrown straight up into the air with an initial velocity of 36 ft/s. its height in feet after t seconds is given by y = 36t - 16t².
find the average velocity for the time period from t = 2 to t = 2 + h, when h is given as follows:
h =.01 sec ft/s
h =.005 sec ft/s
h =.002 sec ft/s
h =.001 sec ft/s
examine your answers above. they should be trending toward the actual value of y(2), which (you can trust us on this, or do the computation yourself using the definition of the derivative), is equal to - 28.
note: you can earn partial credit on this problem.
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Explanation:

Step1: Recall average - velocity formula

The average velocity $v_{avg}$ over the time interval $[a,a + h]$ for a position - function $y(t)$ is given by $v_{avg}=\frac{y(a + h)-y(a)}{h}$. Here, $a = 2$ and $y(t)=36t-16t^{2}$. So, $y(2)=36\times2-16\times2^{2}=72 - 64 = 8$, and $y(2 + h)=36(2 + h)-16(2 + h)^{2}=72+36h-16(4 + 4h+h^{2})=72+36h-64 - 64h-16h^{2}=8 - 28h-16h^{2}$.

Step2: Substitute into average - velocity formula

$v_{avg}=\frac{y(2 + h)-y(2)}{h}=\frac{(8 - 28h-16h^{2})-8}{h}=\frac{-28h-16h^{2}}{h}=-28 - 16h$.

Step3: Calculate for different values of h

When $h = 0.01$

Substitute $h = 0.01$ into $v_{avg}=-28 - 16h$. Then $v_{avg}=-28-16\times0.01=-28 - 0.16=-28.16$ ft/s.

When $h = 0.005$

Substitute $h = 0.005$ into $v_{avg}=-28 - 16h$. Then $v_{avg}=-28-16\times0.005=-28 - 0.08=-28.08$ ft/s.

When $h = 0.002$

Substitute $h = 0.002$ into $v_{avg}=-28 - 16h$. Then $v_{avg}=-28-16\times0.002=-28 - 0.032=-28.032$ ft/s.

When $h = 0.001$

Substitute $h = 0.001$ into $v_{avg}=-28 - 16h$. Then $v_{avg}=-28-16\times0.001=-28 - 0.016=-28.016$ ft/s.

Answer:

-28.16 ft/s
-28.08 ft/s
-28.032 ft/s
-28.016 ft/s