Question

if a rock is thrown from the ground with an initial velocity of 120 ft/sec, then the height of the rock, in feet, at t seconds can be modeled by s(t)=120t - 16t², 0≤t≤7.5. (a) the velocity is v(t)= (b) the time at which the rock reaches its maximum height is t = seconds. (c) the maximum height the rock reaches is feet. (d) the acceleration of the rock when it reaches its maximum height is --select-- hint: follow example 9 and note that the velocity at maximum height is zero.

Explanation:

Step1: Find the velocity function

The velocity function $v(t)$ is the derivative of the position - function $s(t)$. Given $s(t)=120t - 16t^{2}$, by the power rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$, we have $v(t)=\frac{d}{dt}(120t-16t^{2})=120 - 32t$.

Step2: Find the acceleration function

The acceleration function $a(t)$ is the derivative of the velocity function. Since $v(t)=120 - 32t$, then $a(t)=\frac{d}{dt}(120 - 32t)=- 32$.

Step3: Find the time when the rock reaches its maximum height

At the maximum - height, the velocity $v(t)=0$. Set $v(t)=120 - 32t = 0$. Solving for $t$:
\[

$$\begin{align*} 120-32t&=0\\ 32t&=120\\ t&=\frac{120}{32}=\frac{15}{4}=3.75 \end{align*}$$

\]

Step4: Find the maximum height

Substitute $t = 3.75$ into the position function $s(t)$.
\[

$$\begin{align*} s(3.75)&=120\times3.75-16\times(3.75)^{2}\\ &=450-16\times14.0625\\ &=450 - 225\\ &=225 \end{align*}$$

\]

Answer:

(a) $v(t)=120 - 32t$
(b) $t = 3.75$
(c) $225$
(d) $a(t)=-32$