y = x^{2}

\frac{dy}{dx} = 2x

\frac{dv}{dt} = a = −g

v = at + v_{0} = −gt + v_{0}

\frac{dy}{dt} = v = −gt + v_{0}

y = −\frac{1}{2}gt^{2} + v_{0}t + y_{0}

\frac{dv}{dt} = −g + Kv^{2}

\frac{d^{2}y}{dt^{2}} = \frac{dv}{dt} = a = −g = −9.8 m/s^{2}

v(t) = v_{0} − gt

v(t) = 0 − gt

y(t) = y_{0} + v_{0}t − \frac{1}{2}gt^{2}

y(t) = H − \frac{1}{2}gt^{2}

\frac{dv}{dt} = −g

v(t) = v(t_{0}) + v′(t − t_{0}) + \frac{1}{2}v′′(t − t_{0})^{2} + ...

v(t) = v(t_{0}) + v′(t − t_{0})

v_{i+1} = v_{i} + v′_{i}(t_{i+1} − t_{i})

dt = t_{i+1} − t

v′ = \frac{dv}{dt} = −g

v′′ = \frac{d^{2}v}{dt^{2}} = 0

v(t) = v(t_{0}) + (−g)(t − t_{0}) + 0 + 0 ...

v(t) = v(t_{0}) + (−g)(t − t_{0})

v_{i+1} = v_{i} + (−g)(t_{i+1} − t_{i})

y(t) = y(t_{0}) + y′(t − t_{0}) + \frac{1}{2}y′′(t − t_{0})^{2} + ...

\begin{aligned}y′ & = v \\ y′′ & = a = −g \\ y′′′ & = 0 \end{aligned}

y_{i+1} = y_{i} + v(t_{i+1} − t_{i})

error = \frac{1}{2} g(t_{i+1} − t_{i})^{2}

y_{i+1} = y_{i} + v(t_{i+1} − t_{i}) − \frac{1}{2}g(t_{i+1} − t_{i})^{2}