Questions

Let the height of the rock be given by the function $s(t) = 20t - 0.8t^2$ meters, where $t$ is the time in seconds and the initial upward velocity is 20 m/s.

a. Find the rock's velocity and acceleration at time t

The velocity $v(t)$ is the first derivative of the position function $s(t)$ with respect to time $t$. $v(t) = \frac{ds}{dt}$ $v(t) = \frac{d}{dt}(20t - 0.8t^2)$ $v(t) = 20 - 2(0.8)t$ $v(t) = 20 - 1.6t \text{ m/sec}$

The acceleration $a(t)$ is the first derivative of the velocity function $v(t)$ with respect to time $t$, or the second derivative of the position function $s(t)$. $a(t) = \frac{dv}{dt}$ $a(t) = \frac{d}{dt}(20 - 1.6t)$ $a(t) = -1.6 \text{ m/sec}^2$

b. How long does it take the rock to reach its highest point?

The rock reaches its highest point when its vertical velocity is momentarily zero. Set $v(t) = 0$: $20 - 1.6t = 0$ $1.6t = 20$ $t = \frac{20}{1.6}$ $t = 12.5 \text{ sec}$

c. How high does the rock go?

To find the maximum height, substitute the time at which it reaches its highest point ($t = 12.5$ sec) into the position function $s(t)$: $s(12.5) = 20(12.5) - 0.8(12.5)^2$ $s(12.5) = 250 - 0.8(156.25)$ $s(12.5) = 250 - 125$ $s(12.5) = 125 \text{ meters}$

d. How long does it take the rock to reach half its maximum height?

Half of its maximum height is $\frac{125}{2} = 62.5$ meters. Set $s(t) = 62.5$: $20t - 0.8t^2 = 62.5$ Rearrange into a standard quadratic equation $at^2 + bt + c = 0$: $0.8t^2 - 20t + 62.5 = 0$

Use the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: Here, $a = 0.8$, $b = -20$, $c = 62.5$. $t = \frac{-(-20) \pm \sqrt{(-20)^2 - 4(0.8)(62.5)}}{2(0.8)}$ $t = \frac{20 \pm \sqrt{400 - 200}}{1.6}$ $t = \frac{20 \pm \sqrt{200}}{1.6}$ $t = \frac{20 \pm 10\sqrt{2}}{1.6}$ $t = \frac{20 \pm 14.142}{1.6}$

Two possible values for $t$: $t_1 = \frac{20 - 14.142}{1.6} = \frac{5.858}{1.6} \approx 3.661 \text{ sec}$ $t_2 = \frac{20 + 14.142}{1.6} = \frac{34.142}{1.6} \approx 21.339 \text{ sec}$

The question asks how long it takes to reach half its maximum height, implying the first time it reaches that height on the way up. So, we choose the smaller value. It takes approximately 3.661 seconds to reach half its maximum height on the way up.

e. How long is the rock aloft?

The rock is aloft until it returns to the surface, meaning its height $s(t)$ is 0. Set $s(t) = 0$: $20t - 0.8t^2 = 0$ Factor out $t$: $t(20 - 0.8t) = 0$

This gives two solutions: $t = 0$ (the initial launch time) $20 - 0.8t = 0$ $0.8t = 20$ $t = \frac{20}{0.8}$ $t = 25 \text{ sec}$

So, the rock is aloft for 25 seconds.