this is section 2.4 problem 66: if a rock is thrown from the ground with an initial velocity of 120 ft/sec…

Explanation:

Step1: Recall the relationship between position, velocity and acceleration

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

Step2: Recall the relationship between velocity and acceleration

The acceleration function $a(t)$ is the derivative of the velocity function $v(t)$. 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 $s(t)=120t-16t^{2}$. [ \begin{align*} s(3.75)&=120\times3.75-16\times(3.75)^{2}\ &=450-16\times\frac{14.0625}{1}\ &=450 - 225\ &=225 \end{align*} ]

Step5: Find the acceleration at the maximum height

The acceleration $a(t)=-32$ is a constant function. So, when the rock reaches its maximum height ($t = 3.75$), the acceleration $a(3.75)=-32$.

Answer:

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