(1 point)\na ball is thrown straight up into the air with an initial velocity of 36 ft/s. its height in feet…

(1 point)\na 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.\nfind the average velocity for the time period from t = 2 to t = 2 + h, when h is given as follows:\nh =.01 sec ft/s\nh =.005 sec ft/s\nh =.002 sec ft/s\nh =.001 sec ft/s\nexamine 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.\nnote: you can earn partial credit on this problem.\npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have unlimited attempts remaining.\npage generated at 09/20/2025 at 04:22pm pdt\nwebwork © 1996 - 2023 | theme math4 | ww_version: 2.18 | pg_version 2.18\nthe webwork project

(1 point)\na 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.\nfind the average velocity for the time period from t = 2 to t = 2 + h, when h is given as follows:\nh =.01 sec ft/s\nh =.005 sec ft/s\nh =.002 sec ft/s\nh =.001 sec ft/s\nexamine 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.\nnote: you can earn partial credit on this problem.\npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have unlimited attempts remaining.\npage generated at 09/20/2025 at 04:22pm pdt\nwebwork © 1996 - 2023 | theme math4 | ww_version: 2.18 | pg_version 2.18\nthe webwork project

Answer

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