# Homework Help: Projectile subject to air resistance

1. Feb 19, 2012

### rg2004

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

Consider a projectile subject to air resistance. The drag force is $F=-\hat{v} C v^{a}$ where C and a are constants and v=$\dot{r}$. Restrict the problem to 2-D and take y in the vertical direction. Take the magnitude of initial velocity to be $v_{0}$ and the initial position to be the origin. Introduce dimensionless coordinates X,Y,Z by relations

$x=\frac{v_{0}^{2}}{g}X$
$y=\frac{v_{0}^{2}}{g}Y$
$t=\frac{v_{0}}{g}T$

Given the above definitions show that

$\frac{d^2Y}{dT^2}=-1-k V^{a-1} \frac{dY}{dT}$
$\frac{d^2X}{dT^2}=-k V^{a-1} \frac{dX}{dT}$

where

$k=\frac{C v_{0}^a}{m g}$

$V=\sqrt{(\frac{dY}{dT})^2+(\frac{dX}{dT})^2}$

3. The attempt at a solution

$x=\frac{v_{0}^{2}}{g}X$
$dx=\frac{v_{0}^{2}}{g}dX$

$y=\frac{v_{0}^{2}}{g}Y$
$dy=\frac{v_{0}^{2}}{g}dY$

$t=\frac{v_{0}}{g}T$
$dt=\frac{v_{0}}{g}dT$

finding d?/dT equations
$\frac{dx}{dt}=\frac{\frac{v_{0}^{2}}{g}dX}{\frac{v_{0}}{g}dT} =v_0\frac{dX}{dT}$

$\frac{dy}{dt}=\frac{\frac{v_{0}^{2}}{g}dY}{\frac{v_{0}}{g}dT} =v_0\frac{dY}{dT}$

finding $\frac{d^2?}{dT^2}$
$\frac{d^2x}{dt^2}=\frac{dx}{dt}\frac{dx}{dt}=v_0^2\frac{d^2X}{dT^2}$

$\frac{d^2y}{dt^2}=\frac{dy}{dt}\frac{dy}{dt}=v_0^2\frac{d^2Y}{dT^2}$

From force equation:
$F_{d,x}=m \ddot{x}=-\hat{v} C v^{a}=\frac{-\dot{x}}{\sqrt{\dot{x}^2+\dot{y}^2}} C (\sqrt{\dot{x}^2+\dot{y}^2})^{a}$
$=-\dot{x} C \sqrt{\dot{x}^2+\dot{y}^2}^{a-1}$

$m v_0^2\frac{d^2X}{dT^2}=-(v_0\frac{dX}{dT}) C \sqrt{(v_0\frac{dX}{dT})^2+(v_0\frac{dY}{dT})^2}^{a-1}$

$m v_0^2\frac{d^2X}{dT^2}=-v_0^a \frac{dX}{dT} C \sqrt{(\frac{dX}{dT})^2+(\frac{dY}{dT})^2}^{a-1}$

$\frac{d^2X}{dT^2}=-\frac{v_0^a C}{m v_0^2} V^{a-1} \frac{dX}{dT}$

$\frac{d^2X}{dT^2}=-k \frac{g}{v_0^2} V^{a-1} \frac{dX}{dT}$
Which is not what I'm supposed to get. I must be messing up a rule or two somewhere.

Thank you.

Last edited: Feb 19, 2012
2. Feb 19, 2012

### ehild

Do the second derivatives again. They are not the squares of the first derivatives, as you wrote.

ehild

3. Feb 19, 2012

### rg2004

So then would the derivative be

$\frac{d^2x}{dt^2t}=\frac{d}{dt}\frac{dx}{dt}=\frac{d}{dt}(v_0 \frac{dX}{dT})=\frac{d}{dT}(\frac{g}{v_0})(v_0 \frac{dX}{dT})$

$\frac{d^2x}{dt^2}=g\frac{d^2X}{dT^2}$

similarly

$\frac{d^2y}{dt^2}=g\frac{d^2Y}{dT^2}$