Solution 15.4b
From Mechanics
(Difference between revisions)
(New page: <math>\begin{align} & v=0,\quad a=\text{ }-\text{9}\textrm{.}\text{8},\quad s=\text{ 1}\textrm{.}\text{4} \\ & \\ & {{v}^{2}}={{u}^{2}}+2as \\ & {{0}^{2}}={{u}^{2}}+2\times (-9\textrm{...) |
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<math>\begin{align} | <math>\begin{align} | ||
- | & v=0,\quad a=\text{ }-\text{9}\textrm{.}\text{8},\quad s=\text{ 1}\textrm{.}\text{4} \\ | + | & v=0 \text{ m}{{\text{s}}^{-1}} ,\quad a=\text{ }-\text{9}\textrm{.}\text{8} \text{ m}{{\text{s}}^{-2}},\quad s=\text{ 1}\textrm{.}\text{4 m} \\ |
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& {{v}^{2}}={{u}^{2}}+2as \\ | & {{v}^{2}}={{u}^{2}}+2as \\ |
Current revision
\displaystyle \begin{align} & v=0 \text{ m}{{\text{s}}^{-1}} ,\quad a=\text{ }-\text{9}\textrm{.}\text{8} \text{ m}{{\text{s}}^{-2}},\quad s=\text{ 1}\textrm{.}\text{4 m} \\ & \\ & {{v}^{2}}={{u}^{2}}+2as \\ & {{0}^{2}}={{u}^{2}}+2\times (-9\textrm{.}8)\times 1\textrm{.}4 \\ & v=5\textrm{.}24\text{ m}{{\text{s}}^{\text{-1}}} \\ \end{align}