Solve y=1/2gt^2+V_{0}t

$$y = \frac { 1 } { 2 } g t ^ { 2 } + V _ { 0 } t$$

Solve for V_0 (complex solution)

$$\left\{\begin{matrix}V_{0}=-\frac{gt}{2}+\frac{y}{t}\text{, }&t\neq 0\\V_{0}\in \mathrm{C}\text{, }&y=0\text{ and }t=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$\frac{1}{2}gt^{2}+V_{0}t=y$$

Subtract \frac{1}{2}gt^{2} from both sides.

$$V_{0}t=y-\frac{1}{2}gt^{2}$$

The equation is in standard form.

$$tV_{0}=-\frac{gt^{2}}{2}+y$$

Divide both sides by t.

$$\frac{tV_{0}}{t}=\frac{-\frac{gt^{2}}{2}+y}{t}$$

Dividing by t undoes the multiplication by t.

$$V_{0}=\frac{-\frac{gt^{2}}{2}+y}{t}$$

Divide y-\frac{gt^{2}}{2} by t.

$$V_{0}=-\frac{gt}{2}+\frac{y}{t}$$

Solve for g (complex solution)

$$\left\{\begin{matrix}g=-\frac{2\left(V_{0}t-y\right)}{t^{2}}\text{, }&t\neq 0\\g\in \mathrm{C}\text{, }&y=0\text{ and }t=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$\frac{1}{2}gt^{2}+V_{0}t=y$$

Subtract V_{0}t from both sides.

$$\frac{1}{2}gt^{2}=y-V_{0}t$$

The equation is in standard form.

$$\frac{t^{2}}{2}g=y-V_{0}t$$

Divide both sides by \frac{1}{2}t^{2}.

$$\frac{2\times \frac{t^{2}}{2}g}{t^{2}}=\frac{2\left(y-V_{0}t\right)}{t^{2}}$$

Dividing by \frac{1}{2}t^{2} undoes the multiplication by \frac{1}{2}t^{2}.

$$g=\frac{2\left(y-V_{0}t\right)}{t^{2}}$$

Solve for V_0

$$\left\{\begin{matrix}V_{0}=-\frac{gt}{2}+\frac{y}{t}\text{, }&t\neq 0\\V_{0}\in \mathrm{R}\text{, }&y=0\text{ and }t=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$\frac{1}{2}gt^{2}+V_{0}t=y$$

Subtract \frac{1}{2}gt^{2} from both sides.

$$V_{0}t=y-\frac{1}{2}gt^{2}$$

The equation is in standard form.

$$tV_{0}=-\frac{gt^{2}}{2}+y$$

Divide both sides by t.

$$\frac{tV_{0}}{t}=\frac{-\frac{gt^{2}}{2}+y}{t}$$

Dividing by t undoes the multiplication by t.

$$V_{0}=\frac{-\frac{gt^{2}}{2}+y}{t}$$

Divide y-\frac{gt^{2}}{2} by t.

$$V_{0}=-\frac{gt}{2}+\frac{y}{t}$$

Solve for g

$$\left\{\begin{matrix}g=-\frac{2\left(V_{0}t-y\right)}{t^{2}}\text{, }&t\neq 0\\g\in \mathrm{R}\text{, }&y=0\text{ and }t=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$\frac{1}{2}gt^{2}+V_{0}t=y$$

Subtract V_{0}t from both sides.

$$\frac{1}{2}gt^{2}=y-V_{0}t$$

The equation is in standard form.

$$\frac{t^{2}}{2}g=y-V_{0}t$$

Divide both sides by \frac{1}{2}t^{2}.

$$\frac{2\times \frac{t^{2}}{2}g}{t^{2}}=\frac{2\left(y-V_{0}t\right)}{t^{2}}$$

Dividing by \frac{1}{2}t^{2} undoes the multiplication by \frac{1}{2}t^{2}.

$$g=\frac{2\left(y-V_{0}t\right)}{t^{2}}$$