Ian: New page: The energy is constant. Thus the energy in the beginning is equal to the energy at the end. $\begin{align} & \frac{1}{2}\times m\times {{18}^{2}}=\frac{1}{2}\times m\times {{12}^{2}}...$

2010-10-04T11:59:16Z

New page: The energy is constant. Thus the energy in the beginning is equal to the energy at the end. <math>\begin{align} & \frac{1}{2}\times m\times {{18}^{2}}=\frac{1}{2}\times m\times {{12}^{2}}...

New page

The energy is constant. Thus the energy in the beginning is equal to the energy at the end.

<math>\begin{align}
& \frac{1}{2}\times m\times {{18}^{2}}=\frac{1}{2}\times m\times {{12}^{2}}+m\times 9 \textrm{.}8\times h \\
& h=\frac{162-72}{9 \textrm{.}8}=9 \textrm{.}18\text{ m} \\
\end{align}</math>

Solution 17.8a - Revision history

Ian: New page: The energy is constant. Thus the energy in the beginning is equal to the energy at the end. \begin{align} & \frac{1}{2}\times m\times {{18}^{2}}=\frac{1}{2}\times m\times {{12}^{2}}...

Ian: New page: The energy is constant. Thus the energy in the beginning is equal to the energy at the end. $\begin{align} & \frac{1}{2}\times m\times {{18}^{2}}=\frac{1}{2}\times m\times {{12}^{2}}...$