Solution 15.4a
From Mechanics
(Difference between revisions)
(New page: <math>\begin{align} & u=0,\quad a=\text{9}\textrm{.}\text{8},\quad s=\text{2} \\ & {{v}^{2}}={{u}^{2}}+2as \\ & {{v}^{2}}={{0}^{2}}+2\times 9\textrm{.}8\times 2 \\ & v=6\textrm{.}26\tex...) |
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<math>\begin{align} | <math>\begin{align} | ||
- | & u=0,\quad a=\text{9}\textrm{.}\text{8},\quad s=\text{2} \\ | + | & u=0\ \text{ m}{{\text{s}}^{-1}},\quad a=\text{9}\textrm{.}\text{8}\ \text{ m}{{\text{s}}^{-2}},\quad s=\text{2 m} \\ |
& {{v}^{2}}={{u}^{2}}+2as \\ | & {{v}^{2}}={{u}^{2}}+2as \\ | ||
& {{v}^{2}}={{0}^{2}}+2\times 9\textrm{.}8\times 2 \\ | & {{v}^{2}}={{0}^{2}}+2\times 9\textrm{.}8\times 2 \\ | ||
& v=6\textrm{.}26\text{ m}{{\text{s}}^{\text{-1}}} \\ | & v=6\textrm{.}26\text{ m}{{\text{s}}^{\text{-1}}} \\ | ||
\end{align}</math> | \end{align}</math> |
Current revision
\displaystyle \begin{align} & u=0\ \text{ m}{{\text{s}}^{-1}},\quad a=\text{9}\textrm{.}\text{8}\ \text{ m}{{\text{s}}^{-2}},\quad s=\text{2 m} \\ & {{v}^{2}}={{u}^{2}}+2as \\ & {{v}^{2}}={{0}^{2}}+2\times 9\textrm{.}8\times 2 \\ & v=6\textrm{.}26\text{ m}{{\text{s}}^{\text{-1}}} \\ \end{align}