Solve equation [math] v = \frac{s}{t} [/math] by [math]s[/math] and [math]v[/math].[br][br][b]Solution.[/b] To solve the equation by [math]s[/math], we multiply both sides of the equation by [math]t[/math]:[br][br][math] \large \begin{array}{rcll}[br]v & = & \frac{s}{t} & | \cdot t \\[br]v \cdot t & = & s \\[br]s & = & v \cdot t .[br]\end{array} [/math][br][br]To solve the equation by [math]t[/math], we need a few steps:[br][br][math]\large \begin{array}{rcll}[br]v & = & \frac{s}{t} & | \cdot t \\[br]v \cdot t & = & s & | : v \\[br]t & = & \frac{s}{v}[br]\end{array}[/math][br][br]Alternatively, we may solve the equation as a proportion equation, by writing [math]v = \frac{v}{1}[/math]:[br][br][math] \large \begin{array}{rcll}[br]\frac{ \textcolor{red}v}{ \textcolor{blue}1} & = & \frac{ \textcolor{blue}s}{\textcolor{red} t} \\[br]v \cdot t & = & s & | : v \\[br]t & = & \frac{s}{v}[br]\end{array}[/math]

Solve equation [math]E=\frac{1}{2}mv^2[/math] by [math]m[/math] and [math]v[/math].[br][br][b]Solution.[/b] First, let us solve the equation by [math]m[/math]:[br][br][math] \large \begin{array}{rcll}[br]E & = & \frac{1}{2} mv^2 & | \cdot 2 \\[br]2E & = & mv^2 \\[br]mv^2 & = & 2E & | : v^2 \\[br]m & = & \frac{2E}{v^2}[br]\end{array} [/math][br][br]Then we solve the equation by [math]v[/math]:[br][br][math] \large \begin{array}{rcll}[br]E & = & \frac{1}{2} mv^2 & | \cdot 2 \\[br]2E & = & mv^2 \\[br]mv^2 & = & 2E & | : m \\[br]v^2 & = & \frac{2E}{m} & | \sqrt{} \\[br]v & = & \sqrt{\frac{2E}{m}}[br]\end{array} [/math]

The height of a baseball during a throwing motion is given by the equation[br][math]\large h = h_0 + v_0 t - \frac{1}{2} gt^2 . [/math][br]Here [math]h[/math] is the height during a given time [math]t[/math], while [math]h_0[/math] is the initial height (i.e. height at the beginning), [math]v_0[/math] is the initial velocity, and [math]g[/math] is the gravitational acceleration, [math]g = 9,81 \, \mathrm{m/s^2}[/math].[br][br]Let [math]h_0 = 0 \, \mathrm{m}[/math] and [math]v_0 = 10,0 \, \mathrm{m/s}[/math]. We calculate when the baseball hits the ground. At ground level, [math]h=0 \, \mathrm{m}[/math], so we can write:[br][br][math] \large \begin{array}{rcll}[br]h_0 + v_0 t - \frac{1}{2}gt^2 & = & 0 & | \ h_0 = 0 \\[br]v_0 t - \frac{1}{2}gt^2 & = & 0 & | \cdot 2 \\[br]2 v_0 t - gt^2 & = & 0 \\[br]t(2 v_0 -gt) & = & 0 \\[br]\end{array} [/math][br][br]So according to the [b]zero product property[/b], either [math]t = 0[/math] (this is the time when the baseball is thrown in the air), or[br][br][math] \large \begin{array}{rcll}[br]2v_0 - gt & = & 0 \\[br]2v_0 & = & gt \\[br]gt & = & 2v_0 & | : g \\[br]t & = & \frac{2 v_0}{g}[br]\end{array} [/math][br][br]Now we can calculate the value for [math]t[/math]:[br][br][math] \large t = \frac{2 \cdot 10,0 \, \mathrm{m/s}}{9,81 \, \mathrm{m/s^2}} \approx 2,04 \, \mathrm{s} . [/math]