1. The problem statement, all variables and given/known data

A skydiver drops o¤ an outcrop at the top of a sheer face on a mountain and falls vertically
downwards. Let $v(x)$ be the velocity of the skydiver at a vertical distance $x$ below the drop-off point. As the skydiver falls he will experience a drag force of magnitude
$\beta v^2$ where ## \beta ## is a positive constant. Make use of Newton’s Second Law together with the Chain Rule to show that $$m \frac{dv}{dx} v = mg – \beta v^2$$ where m is his mass and g is the acceleration due to gravity. Explain why $$v(o) = 0$$

show that $$\int \frac{v}{v_t ^2 – v^2}\,dv = \frac{k}{m} x + c$$

where $$v_t = \sqrt{ \frac{m}{\beta} g }$$

2. Relevant equations

$\sum \vec{F} = m\vec{a}$

3. The attempt at a solution

$\sum \vec{F} = m\vec{a} = mg – \beta v^2 = ma$

$m \frac{dv}{dt} = mg – \beta v^2$

$m \frac{dv}{dx} \frac{dx}{dt} = mg – \beta v^2$

$m \frac{dv}{dx} v = mg – \beta v^2$

$v(0) = 0$. Since at position $x = 0$ (start position) the skydiver has not jumped so his starting velocity ## (v) ## is ## 0 \frac{m}{s} ##. Meaning that the slope of the velocity is ## g ##.

$\int \frac{v}{v_t ^2 – v^2}\,dv = \frac{\beta}{m} x + c$
Letting ## u = v_t ^2 – v^2 ##, we have ## \frac{du}{dv} = -2v \Rightarrow vdv = – \frac{1}{2} du ##.

$\int \frac{v}{v_t ^2 – v^2}\,dv$

$= -\frac{1}{2} \int \frac{1}{u}\,du$

$= -\frac{1}{2} ( ln(1) – ln(v_t ^2 – v^2) ) + c$

$= -\frac{1}{2} ln(v_t ^2 – v^2) +c$

$= ln \Big ( \frac{1}{\sqrt {v_t ^2 – v^2}} \Big)$

I’m not entirely sure where to go from here. How do I get an ## x ## on the right side of the expression above from this integration.

http://ift.tt/1jui6nb