Solution 15.4b

From Mechanics

(Difference between revisions)
Jump to: navigation, search
(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{...)
Current revision (18:26, 14 September 2010) (edit) (undo)
 
Line 1: Line 1:
<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} \\
& \\
& \\
& {{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}