Single Variable Calculus, Chapter 3, 3.7, Section 3.7, Problem 9

Suppose that a stone is thrown vertically upward from the surface of the moon with a velocity of $\displaystyle \frac{10m}{s}$, its height(in meters) after $t$ seconds is $h = 10t - 0.83t^2$

a.) What is the velocity of the stone after $3s$?
b.) What is the velocity of the stone after it has risen $25m$?


$
\begin{equation}
\begin{aligned}
\text{velocity } = h'(t) = \frac{dh}{dt} &\\
\\
\frac{dh}{dt} &= 10\frac{d}{dt} (t) - 0.83 \frac{d}{dt} (t^2)\\
\\
\frac{dh}{dt} &= 10 - 1.66 t
\end{aligned}
\end{equation}
$


when $t = 3s$,

$
\begin{equation}
\begin{aligned}
\text{velocity } &= 10 - 1.66(3)\\
&= 5.02 \frac{m}{s}
\end{aligned}
\end{equation}
$


b.) @ $h = 25m$,
$25 = 10t - 0.83t^2$
$ 0.83t^2 - 10t + 25 = 0$
Using Quadratic formula,
$t = 3.54$ and $ t = 8.51$

The velocity at $ t = 3.54s$
$\displaystyle \nu(t) = 10 - 1.66(3.54) = 4.12 \frac{m}{s}$
The velocity at $t =8.51s$
$\displaystyle \nu(t) = 10 - 1.66(8.51) = -4.12 \frac{m}{s}$

Assuming that the required velocity is the upward velocity at $\displaystyle 4.12\frac{m}{s}$