**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 [itex] v(x) [/itex] be the velocity of the skydiver at a vertical distance [itex] x [/itex] below the drop-off point. As the skydiver falls he will experience a drag force of magnitude

[itex] \beta v^2 [/itex] 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 [tex] v(o) = 0 [/tex]

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

where [tex] v_t = \sqrt{ \frac{m}{\beta} g } [/tex]

**2. Relevant equations**

[itex] \sum \vec{F} = m\vec{a} [/itex]

**3. The attempt at a solution**

[itex] \sum \vec{F} = m\vec{a} = mg – \beta v^2 = ma [/itex]

[itex] m \frac{dv}{dt} = mg – \beta v^2 [/itex]

[itex] m \frac{dv}{dx} \frac{dx}{dt} = mg – \beta v^2 [/itex]

[itex] m \frac{dv}{dx} v = mg – \beta v^2 [/itex]

[itex] v(0) = 0 [/itex]. Since at position [itex] x = 0 [/itex] (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 ##.

[itex] \int \frac{v}{v_t ^2 – v^2}\,dv = \frac{\beta}{m} x + c [/itex]

Letting ## u = v_t ^2 – v^2 ##, we have ## \frac{du}{dv} = -2v \Rightarrow vdv = – \frac{1}{2} du ##.

[itex] \int \frac{v}{v_t ^2 – v^2}\,dv [/itex]

[itex] = -\frac{1}{2} \int \frac{1}{u}\,du [/itex]

[itex] = -\frac{1}{2} ( ln(1) – ln(v_t ^2 – v^2) ) + c [/itex]

[itex] = -\frac{1}{2} ln(v_t ^2 – v^2) +c [/itex]

[itex] = ln \Big ( \frac{1}{\sqrt {v_t ^2 – v^2}} \Big) [/itex]

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