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 \((n=0)\) :

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