a rock thrown vertically upward from the surface of the moon at a velocity of 28 m/sec reaches a height of s…

Explanation:

Step1: Recall the relationship between position, velocity and acceleration

Velocity $v(t)$ is the derivative of position $s(t)$, and acceleration $a(t)$ is the derivative of velocity $v(t)$. Given $s(t)=28t - 0.8t^{2}$.

Step2: Differentiate $s(t)$ to find $v(t)$

Using the power - rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$, we have $v(t)=\frac{d}{dt}(28t - 0.8t^{2})=28-1.6t$.

Step3: Differentiate $v(t)$ to find $a(t)$

Differentiating $v(t)=28 - 1.6t$ with respect to $t$, we get $a(t)=\frac{d}{dt}(28 - 1.6t)=- 1.6$.

Step4: Find the time to reach the highest point

At the highest point, $v(t)=0$. Set $v(t)=28 - 1.6t = 0$. Solving for $t$ gives $t=\frac{28}{1.6}=17.5$ s.

Step5: Find the maximum height

Substitute $t = 17.5$ into $s(t)$. So $s(17.5)=28\times17.5-0.8\times(17.5)^{2}=490 - 0.8\times306.25=490 - 245=245$ m.

Step6: Find the time to reach half - maximum height

Set $s(t)=\frac{245}{2}=122.5$. So $28t-0.8t^{2}=122.5$. Rearranging gives $0.8t^{2}-28t + 122.5 = 0$. Using the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$. Here $a = 0.8$, $b=-28$, $c = 122.5$. First, calculate the discriminant $\Delta=b^{2}-4ac=(-28)^{2}-4\times0.8\times122.5=784 - 392 = 392$. Then $t=\frac{28\pm\sqrt{392}}{1.6}=\frac{28\pm19.8}{1.6}$. We get $t_1=\frac{28 + 19.8}{1.6}\approx29.9$ s and $t_2=\frac{28 - 19.8}{1.6}\approx5.1$ s.

Step7: Find the time the rock is aloft

Set $s(t)=0$. So $28t-0.8t^{2}=0$. Factoring out $t$ gives $t(28 - 0.8t)=0$. We have $t = 0$ (corresponds to the time of throwing) and $28-0.8t=0$, which gives $t=\frac{28}{0.8}=35$ s.

Answer:

a. $v = 28-1.6t$ m/s, $a=-1.6$ m/s² b. $17.5$ s c. $245$ m d. Approximately $5.1$ s and $29.9$ s e. $35$ s