Problem source

A particle of unit mass is projected vertically upwards with speed $u$. At height $x$, while the particle is moving upwards, it is found to experience a total force $F$, due to gravity and air resistance, given by $F=\alpha \e^{-\beta x}$, where $\alpha$ and $\beta$ are positive constants. Calculate the energy expended in reaching this height. Show that 
\[
F= {\textstyle \frac12} \beta v^2+ \alpha - {\textstyle \frac12} \beta u^2 \;,
\]
where $v$ is the speed of the particle, and explain why $ \alpha = \frac12 \beta u^2 +g$, where $g$ is the acceleration due to gravity. 
Determine  an expression, in terms of $y$, $g$ and $\beta$, for the air resistance  experienced by the particle on its downward journey when it is at a distance $y$ below its highest point.

Solution source

Considering the energy of the particle, we have initial kinetic energy of $\frac12 u^2$ and final energy is $\frac12 v^2$, the change in energy is the work done by the force,

\begin{align*}
    &&\text{Work done against resistance} &= \text{loss in kinetic energy} \\
    &&\int F \, \d x &= \int \alpha e^{-\beta x} \, \d x \\
    &&&= \frac{\alpha}{\beta} \l 1 - e^{-\beta x} \r \\
    &&&= \frac{1}{\beta} \l \alpha - F\r \\
    &&&= \frac12 u^2 - \frac12 v^2 \\
    \Rightarrow && F &= \frac12 \beta v^2 + \alpha - \frac12 \beta u^2
\end{align*}  

When $v = 0$ there is no air resistance, ie $F = g$, but $g = 0 + \alpha - \frac12 \beta u^2 \Rightarrow \alpha = g + \frac12 \beta u^2$

$F = \frac12 \beta v^2 + g$, ie air resistance is $\frac12 \beta v^2$

Looking at forces acting on the particle when it's descending,

\begin{align*}
    && v \frac{dv}{dx} &= g - \frac12 \beta v^2 \\
    \Rightarrow && \frac{v}{g - \frac12 \beta v^2} \frac{dv}{dx} &= 1 \\
    \Rightarrow && \int \frac{v}{g - \frac12 \beta v^2} \, dv &= \int dx \\
    \Rightarrow &&  \frac1{\beta}\l\ln(g - \frac12\beta v^2) - \ln(g)\r &= y\\
    \Rightarrow && \ln \l 1 - \frac12 \frac{\beta}{g}v^2 \r &= \beta y \\
    \Rightarrow && \frac{g}{\beta} \l 1-e^{-\beta y} \r = \frac12 v^2
\end{align*}

Since force is the rate of change of work, we can say that the force is $ge^{-\beta y}$ and the air resistance is $g \l 1-e^{-\beta y} \r$