Réitigh h=1/2gt^2+u_{0}t

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

Réitigh do g.

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

Céimeanna chun Cothromóidí Líneacha a Réiteach

Athraigh na taobhanna ionas go mbeidh na téarmaí inathraitheacha ar fad ar an taobh clé.

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

Bain u_{0}t ón dá thaobh.

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

Tá an chothromóid i bhfoirm chaighdeánach.

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

Roinn an dá thaobh faoi \frac{1}{2}t^{2}.

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

Má roinntear é faoi \frac{1}{2}t^{2} cuirtear an iolrúchán faoi \frac{1}{2}t^{2} ar ceal.

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

Réitigh do h.

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