Solve y=40+V_{0}t-1/2gt^2

$$y = 40 + V _ { 0 } t - \frac { 1 } { 2 } g t ^ { 2 }$$

Solve for V_0 (complex solution)

$$\left\{\begin{matrix}V_{0}=\frac{2y+gt^{2}-80}{2t}\text{, }&t\neq 0\\V_{0}\in \mathrm{C}\text{, }&y=40\text{ and }t=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$40+V_{0}t-\frac{1}{2}gt^{2}=y$$

Add \frac{1}{2}gt^{2} to both sides.

$$40+V_{0}t=y+\frac{1}{2}gt^{2}$$

Subtract 40 from both sides.

$$V_{0}t=y+\frac{1}{2}gt^{2}-40$$

The equation is in standard form.

$$tV_{0}=\frac{gt^{2}}{2}+y-40$$

Divide both sides by t.

$$\frac{tV_{0}}{t}=\frac{\frac{gt^{2}}{2}+y-40}{t}$$

Dividing by t undoes the multiplication by t.

$$V_{0}=\frac{\frac{gt^{2}}{2}+y-40}{t}$$

Divide y-40+\frac{gt^{2}}{2} by t.

$$V_{0}=\frac{gt}{2}+\frac{y-40}{t}$$

Solve for g (complex solution)

$$\left\{\begin{matrix}g=\frac{2\left(40+V_{0}t-y\right)}{t^{2}}\text{, }&t\neq 0\\g\in \mathrm{C}\text{, }&y=40\text{ and }t=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$40+V_{0}t-\frac{1}{2}gt^{2}=y$$

Subtract 40 from both sides.

$$V_{0}t-\frac{1}{2}gt^{2}=y-40$$

Subtract V_{0}t from both sides.

$$-\frac{1}{2}gt^{2}=y-40-V_{0}t$$

The equation is in standard form.

$$\left(-\frac{t^{2}}{2}\right)g=y-V_{0}t-40$$

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

$$\frac{\left(-\frac{t^{2}}{2}\right)g}{-\frac{t^{2}}{2}}=\frac{y-V_{0}t-40}{-\frac{t^{2}}{2}}$$

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

$$g=\frac{y-V_{0}t-40}{-\frac{t^{2}}{2}}$$

Divide y-40-V_{0}t by -\frac{1}{2}t^{2}.

$$g=-\frac{2\left(y-V_{0}t-40\right)}{t^{2}}$$

Solve for V_0

$$\left\{\begin{matrix}V_{0}=\frac{2y+gt^{2}-80}{2t}\text{, }&t\neq 0\\V_{0}\in \mathrm{R}\text{, }&y=40\text{ and }t=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$40+V_{0}t-\frac{1}{2}gt^{2}=y$$

Add \frac{1}{2}gt^{2} to both sides.

$$40+V_{0}t=y+\frac{1}{2}gt^{2}$$

Subtract 40 from both sides.

$$V_{0}t=y+\frac{1}{2}gt^{2}-40$$

The equation is in standard form.

$$tV_{0}=\frac{gt^{2}}{2}+y-40$$

Divide both sides by t.

$$\frac{tV_{0}}{t}=\frac{\frac{gt^{2}}{2}+y-40}{t}$$

Dividing by t undoes the multiplication by t.

$$V_{0}=\frac{\frac{gt^{2}}{2}+y-40}{t}$$

Divide y-40+\frac{gt^{2}}{2} by t.

$$V_{0}=\frac{gt}{2}+\frac{y-40}{t}$$

Solve for g

$$\left\{\begin{matrix}g=\frac{2\left(40+V_{0}t-y\right)}{t^{2}}\text{, }&t\neq 0\\g\in \mathrm{R}\text{, }&y=40\text{ and }t=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$40+V_{0}t-\frac{1}{2}gt^{2}=y$$

Subtract 40 from both sides.

$$V_{0}t-\frac{1}{2}gt^{2}=y-40$$

Subtract V_{0}t from both sides.

$$-\frac{1}{2}gt^{2}=y-40-V_{0}t$$

The equation is in standard form.

$$\left(-\frac{t^{2}}{2}\right)g=y-V_{0}t-40$$

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

$$\frac{\left(-\frac{t^{2}}{2}\right)g}{-\frac{t^{2}}{2}}=\frac{y-V_{0}t-40}{-\frac{t^{2}}{2}}$$

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

$$g=\frac{y-V_{0}t-40}{-\frac{t^{2}}{2}}$$

Divide y-40-V_{0}t by -\frac{1}{2}t^{2}.

$$g=-\frac{2\left(y-V_{0}t-40\right)}{t^{2}}$$