Problem 1

JT 29 Nov 2006 20:47

Let z denote the distance fallen so that the height at time t is h - z(t). The equation of motion is

(1)\begin{align} F=mg-\gamma v=m\frac{d^2z}{dt^2} \end{align}

The terminal velocity is attained when the net force on the body is zero. This gives

(2)\begin{align} v_{\infty}=\frac{mg}{\gamma} \end{align}

To solve the equation of motion we use the result that

(3)\begin{align} \frac{d^2z}{dt^2}=v\frac{dv}{dz} \end{align}

Integrating gives

(4)\begin{align} z=-\frac{m^2g}{\gamma^2}\ln(1-v/v_{\infty}) \end{align}

Integrating again gives

(5)\begin{align} z=\frac{m^2g}{\gamma^2}\ln(\exp(\frac{\gamma}{m}t)+1) \end{align}

For large t, this gives $z=v_{\infty}t$