Question

a weight is thrown down at a velocity of 15 ft/s from a height of 580 feet. its height is given by (h(t)=-16t^{2}-15t + 580) where (t) is given in seconds.
a. what is the weights average velocity over the interval (0,3)?
the weights average velocity over (0,3) is

b. what is the weights instantaneous velocity at time (t = 2)?
the weights instantaneous velocity at (t = 2) is

Explanation:

Step1: Recall average - velocity formula

The average - velocity formula over the interval $[a,b]$ for a position function $h(t)$ is $v_{avg}=\frac{h(b)-h(a)}{b - a}$. Here, $a = 0$, $b = 3$, and $h(t)=-16t^{2}-15t + 580$.

Step2: Calculate $h(3)$ and $h(0)$

First, find $h(3)$:
\[

$$\begin{align*} h(3)&=-16\times3^{2}-15\times3 + 580\\ &=-16\times9-45 + 580\\ &=-144-45 + 580\\ &=-189 + 580\\ &=391 \end{align*}$$

\]
Then, find $h(0)$: $h(0)=-16\times0^{2}-15\times0 + 580 = 580$.

Step3: Calculate average velocity

\[

$$\begin{align*} v_{avg}&=\frac{h(3)-h(0)}{3 - 0}\\ &=\frac{391 - 580}{3}\\ &=\frac{-189}{3}\\ &=-63 \end{align*}$$

\]

Step4: Recall instantaneous - velocity formula

The instantaneous - velocity function $v(t)$ is the derivative of the position function $h(t)$. Using the power rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$, if $h(t)=-16t^{2}-15t + 580$, then $v(t)=h^\prime(t)=-32t-15$.

Step5: Calculate instantaneous velocity at $t = 2$

Substitute $t = 2$ into $v(t)$:
\[

$$\begin{align*} v(2)&=-32\times2-15\\ &=-64-15\\ &=-79 \end{align*}$$

\]

Answer:

a. $-63$
b. $-79$