A particle of unit mass is projected vertically upwards with speed u. At any 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 a given height z. Show that
$F=\frac 1 2 \beta v^2 + \alpha - \frac 1 2 \beta u^2$
where $v$ is the speed of the particle, and explain why $\alpha = \frac 1 2 \beta u^2 + g$ 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.
Fairly long question but I can solve it for the most part.
The problem I have is the last part.
The model solution requires I use $v\frac {dv} {dy}= g - \frac 1 2 \beta v^2$ (since the air res. is going the other way) and it leads to $\frac 1 2 \beta v^2 = g(1-e^{-\beta y})$.
When I was solving it I tried to use conservation of energy, $mgy = \frac 1 2 v^2$, where $m$ = 1. Which lead to $\frac 1 2 \beta v^2 = \beta g y$. Why is this wrong?
Many thanks in advance,
Chris
