แก้โจทย์ z_{1}+v_1^2/2=z_{2}+v_2^2/2

$$z _ { 1 } + \frac { v _ { 1 } ^ { 2 } } { 2 } = z _ { 2 } + \frac { v _ { 2 } ^ { 2 } } { 2 }$$

หาค่า v_1

$$v_{1}=\sqrt{v_{2}^{2}+z_{2}-z_{1}}$$

$$v_{1}=-\sqrt{v_{2}^{2}+z_{2}-z_{1}}\text{, }z_{1}\leq z_{2}+v_{2}^{2}$$

หาค่า v_2

$$v_{2}=\sqrt{z_{1}-z_{2}+v_{1}^{2}}$$

$$v_{2}=-\sqrt{z_{1}-z_{2}+v_{1}^{2}}\text{, }z_{1}\geq z_{2}-v_{1}^{2}$$