a rock thrown vertically upward from the surface of the moon at a velocity of 8 m/sec reaches a height of s…

Explanation:

Step1: Recall the height - velocity - acceleration relationships

The height function is given by $s(t)=8t - 0.8t^{2}$. The velocity function $v(t)$ is the derivative of the height function, and the acceleration function $a(t)$ is the derivative of the velocity function. The derivative of $s(t)$ with respect to $t$ using the power rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$: $v(t)=\frac{ds}{dt}=\frac{d}{dt}(8t-0.8t^{2})=8 - 1.6t$ The derivative of $v(t)$ with respect to $t$: $a(t)=\frac{dv}{dt}=\frac{d}{dt}(8 - 1.6t)=- 1.6$

Step2: Find the time to reach the highest - point

At the highest - point, the velocity $v(t)=0$. Set $v(t)=8 - 1.6t = 0$ Solve for $t$: $1.6t=8$ $t=\frac{8}{1.6}=5$ s

Step3: Find the maximum height

Substitute $t = 5$ into the height function $s(t)$: $s(5)=8\times5-0.8\times5^{2}=40 - 0.8\times25=40 - 20=20$ m

Step4: Find the time to reach half of the maximum height

The maximum height is $20$ m, so half of the maximum height is $s = 10$ m. Set $s(t)=8t - 0.8t^{2}=10$ Rearrange to get a quadratic equation: $0.8t^{2}-8t + 10 = 0$ Multiply through by $10$ to clear the decimal: $8t^{2}-80t + 100 = 0$ Divide through by $4$: $2t^{2}-20t + 25 = 0$ Use the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$. Here, $a = 2$, $b=-20$, and $c = 25$. $t=\frac{20\pm\sqrt{(-20)^{2}-4\times2\times25}}{2\times2}=\frac{20\pm\sqrt{400 - 200}}{4}=\frac{20\pm\sqrt{200}}{4}=\frac{20\pm10\sqrt{2}}{4}=\frac{10\pm5\sqrt{2}}{2}$ $t_1=\frac{10 + 5\sqrt{2}}{2}\approx\frac{10+5\times1.414}{2}=\frac{10 + 7.07}{2}=8.54$ s $t_2=\frac{10 - 5\sqrt{2}}{2}\approx\frac{10-7.07}{2}=1.46$ s

Answer:

b. $5$ s c. $20$ m d. $1.46$ s, $8.54$ s