Projectile subject to air resistance

[rg2004]

Homework Statement

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}

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}<br /> =v_0\frac{dX}{dT}<br />

\frac{dy}{dt}=\frac{\frac{v_{0}^{2}}{g}dY}{\frac{v_{0}}{g}dT}<br /> =v_0\frac{dY}{dT}<br />

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.

 

[ehild]

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

\frac{dy}{dt}=\frac{\frac{v_{0}^{2}}{g}dY}{\frac{v_{0}}{g}dT}<br /> =v_0\frac{dY}{dT}<br />

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}

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

ehild

 

[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}

Thanks for your help