Drag Force - Air Resistance Question

[patrickmoloney]

Homework Statement

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

Homework Equations

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

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.

 

[Tanya Sharma]

I think you have put in far more effort than required .

You already have $mv \frac{dv}{dx} = mg - \beta v^2$

Rearranging , $\frac{mvdv}{mg - \beta v^2} = dx$

Take out $\beta$ from the denominator in the LHS ,you get $\frac{m}{\beta}\frac{vdv}{\frac{mg}{\beta} - v^2} = dx$

$\frac{vdv}{\frac{mg}{\beta} - v^2} = \frac{\beta}{m}dx$

Put $v_t = \frac{mg}{\beta}$ in the denominator of LHS .Now integrate both sides of the equation. On the LHS you keep the expression under the integral sign as it is whereas on the RHS you perform integration.