Answer
a) $v(t)=24-1.6t(m/s)$ and $a(t)=-1.6(m/s^2)$
b) After $15$ seconds, the rock reaches its highest point.
c) The rock goes as high as $180m$.
d) It takes the rock about $4.393$ seconds to reach half its maximum height, which is $90m$.
e) The rock is aloft for $30$ seconds.
Work Step by Step
$$s(t)=24t-0.8t^2$$
a) The rock's velocity and acceleration at time $t$ are:
$$v(t)=\frac{ds}{dt}=\frac{d}{dt}(24t-0.8t^2)=24-1.6t(m/s)$$
$$a(t)=\frac{dv}{dt}=\frac{d}{dt}(24-1.6t)=-1.6(m/s^2)$$
b) The rock reaches its highest point when its velocity has decreased to $0$. So finding $t$ so that $v(t)=0$ and we can figure out how long it takes the rock to reach the highest point.
$$v(t)=0$$ $$24-1.6t=0$$ $$t=15(sec)$$
So after $15$ seconds, the rock reaches its highest point.
c) At $t=15$, we have $$s(15)=24\times15-0.8\times15^2=180(m)$$
The rock goes as high as $180m$.
d) Since the maximum height is $180m$, half the maximum height is $90m$.
To find how long it takes the rock to reach $90m$, we find $t$ such that $s(t)=90$: $$24t-0.8t^2=90$$ $$0.8t^2-24t+90=0$$ $$t\approx4.393sec\hspace{1cm}\text{or}\hspace{1cm}t\approx25.607sec$$
We take only the smaller value, because the larger value is the time it takes after already reaching its maximum height and now is falling down back towards the ground.
Therefore, it takes the rock about $4.393$ seconds to reach half its maximum height.
e) To find how long the rock is aloft, we find 2 values of $t$, one the time it leaves the ground and the other the time it falls back to the ground, both having $s(t)=0$.
$$s(t)=0$$ $$24t-0.8t^2=0$$ $$t(24-0.8t)=0$$ $$t=0sec\hspace{1cm}\text{or}\hspace{1cm}t=30sec$$
$t=0$ corresponds to the moment it was thrown into the air.
$t=30$ corresponds to the moment it fell back into the surface of the moon.
Therefore, the rock is aloft for $30$ seconds.
