Need to calculate how many Newtons of force a rocket gives off? Use this algorithm!

$${\mathtt{y1}} = {\frac{{\mathtt{\,-\,}}{\mathtt{m}}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{k}}\right)}}{\mathtt{\,\times\,}}{In}{\left({\frac{\left({\mathtt{T}}{\mathtt{\,-\,}}{\mathtt{mg}}{\mathtt{\,-\,}}{{\mathtt{kv}}}^{{\mathtt{2}}}\right)}{\left({\mathtt{T}}{\mathtt{\,-\,}}{\mathtt{mg}}\right)}}\right)}$$

Use this algorithm to calculate how many Newtons of force a rocket gives off.

m = mass of rocket

g = 9.81 m/s²

v = velocity of rocket

y = height rocket reaches

k = wind resistance forces

T = motor thrust in Newtons (AKA what we're solving for.)

$$\small{\text{$T=?$}}$$

$$\small{\text{$ \begin{array}{rcl} y_1 &=& -\dfrac{m}{2k}\cdot \ln{ \left( \dfrac{T-(mg+kv^2) }{T-mg} \right) }\\\\ -2y_1\frac{k}{m} &=& \ln{ \left( \dfrac{T-(mg+kv^2) }{T-mg} \right) } \qquad | \qquad e^x \\\\ e^{-2y_1\frac{k}{m}} &=& \dfrac{T-(mg+kv^2) }{T-mg} \\\\ \dfrac{1}{e^{2y_1\frac{k}{m}}} &=& \dfrac{T-(mg+kv^2) }{T-mg} \\\\ e^{2y_1\frac{k}{m}} &=& \dfrac{T-mg}{T-(mg+kv^2) } \\\\ \left[ T-(mg+kv^2) \right] ( e^{2y_1\frac{k}{m}} ) &=& T-mg \\\\ T\cdot ( e^{2y_1\frac{k}{m}} )- (mg+kv^2)\cdot ( e^{2y_1\frac{k}{m}} ) &=& T-mg \\\\ T\cdot ( e^{2y_1\frac{k}{m}} - 1 ) &=& (mg+kv^2)\cdot ( e^{2y_1\frac{k}{m}} ) -mg \\\\ \mathbf{T} & \mathbf{=} & \mathbf{\dfrac{(mg+kv^2)\cdot ( e^{2y_1\frac{k}{m}} ) -mg } {( e^{2y_1\frac{k}{m}} - 1 )}} \\\\ \end{array} $}}$$