Solution 17.8a

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

\displaystyle \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}