66.20: Motion of a Falling Body

Given a body of mass $m$ released from rest at time $t=0$ from a height $h$ above the floor, we find the following results about the motion. Here the $y$ axis points upward and has its origin at the floor, and so the acceleration due to gravity is $-g$.

• Position $y$ at time $t$ :

$$y(t)=h-\frac{1}{2} g t^{2} \notag$$

• Velocity $v$ :

$$\begin{aligned} v(t) & =-g t \\ v(y) & =-\sqrt{2 g(h-y)} \end{aligned} \notag$$

• Fall time $t_{f}$ :

$$t_{f}=\sqrt{\frac{2 h}{g}} \notag$$

• Impact velocity $v_{f}$ :

$$v_{f}=-\sqrt{2 g h} \notag$$

Total energy $E$ :

$$E=m g h \notag$$

• Kinetic energy $K$ :

$$\begin{aligned} K(t) & =\frac{1}{2} m g^{2} t^{2} \\ K(y) & =m g(h-y) \end{aligned} \notag$$

• Potential energy $U$ :

$$\begin{aligned} U(t) & =m g h-\frac{1}{2} m g^{2} t^{2} \\ U(y) & =m g y \end{aligned} \notag$$

• Time derivatives of kinetic energy:

$$\begin{aligned} \frac{d K}{d t} & =-m g v=m g^{2} t=m g \sqrt{2 g(h-y)} \\ \frac{d^{2} K}{d t^{2}} & =m g^{2} \end{aligned}\notag$$

• Time derivatives of potential energy:

$$\begin{aligned} \frac{d U}{d t} & =m g v=-m g^{2} t=-m g \sqrt{2 g(h-y)} \\ \frac{d^{2} U}{d t^{2}} & =-m g^{2} \end{aligned} \notag$$

• Time averages:

$$\begin{aligned} \langle v\rangle & =-\frac{1}{2} \sqrt{2 g h} \\ \langle K\rangle & =\frac{1}{3} m g h \\ \langle U\rangle & =\frac{2}{3} m g h \end{aligned} \notag$$

• Virial theorem $(n=0)$ :

$$\langle K\rangle=\frac{1}{2}\langle U\rangle \notag$$