Solve h=1/2gt^2+u_{0}t

$$h = \frac { 1 } { 2 } g t ^ { 2 } + u _ { 0 } t$$

Solve for g

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

Steps for Solving Linear Equation

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

$$\frac{1}{2}gt^{2}+u_{0}t=h$$

Subtract u_{0}t from both sides.

$$\frac{1}{2}gt^{2}=h-u_{0}t$$

The equation is in standard form.

$$\frac{t^{2}}{2}g=h-tu_{0}$$

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

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

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

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

Solve for h

$$h=\frac{t\left(gt+2u_{0}\right)}{2}$$