Here's how the equations of motion are derived:

For a falling particle the gravitational acceleration is given by $$\ddot{y}= g$$ where the 'double dot' denotes differentiation with respect to time $t$ and $y$ is measured vertically downwards. By integrating this equation we can derive the equations of motion where the velocity at time t is denoted by $v = \dot{y}$ and the initial velocity $v(0)=u$. From these equations we can deduce the energy equation.

$$\ddot{y} = g$$ where $y$ is measured upwards. Integrating wrt $t$ $$\dot{y}= v = u + gt $$ Integrating a second time and taking $y = 0$ when $t=0$ $$y = ut + {1\over 2}gt^2$$ Eliminating $t={v - u\over g}$ we get $$y = {u(v - u)\over g} + {g\over 2} {(v - u)^2\over g^2}$$ which simplifies to $$2gy = v^2 - u^2.$$ This is equivalent to the energy equation for a mass $m$ where the change in potential energy in falling a distance $y$ is equal to the change in kinetic energy given by the equation: $$mgy = {1\over 2}mv^2 - {1\over 2}mu^2$$

For a falling particle the gravitational acceleration is given by $$\ddot{y}= g$$ where the 'double dot' denotes differentiation with respect to time $t$ and $y$ is measured vertically downwards. By integrating this equation we can derive the equations of motion where the velocity at time t is denoted by $v = \dot{y}$ and the initial velocity $v(0)=u$. From these equations we can deduce the energy equation.

$$\ddot{y} = g$$ where $y$ is measured upwards. Integrating wrt $t$ $$\dot{y}= v = u + gt $$ Integrating a second time and taking $y = 0$ when $t=0$ $$y = ut + {1\over 2}gt^2$$ Eliminating $t={v - u\over g}$ we get $$y = {u(v - u)\over g} + {g\over 2} {(v - u)^2\over g^2}$$ which simplifies to $$2gy = v^2 - u^2.$$ This is equivalent to the energy equation for a mass $m$ where the change in potential energy in falling a distance $y$ is equal to the change in kinetic energy given by the equation: $$mgy = {1\over 2}mv^2 - {1\over 2}mu^2$$