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Free Fall

An object in free fall (ignoring air resistance) accelerates downwards at a constant rate due to gravity (g).
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The statement of the theorem

Let r⃗(t)=⟨x(t),y(t),z(t)⟩∈R3\vec{r}(t) = \langle x(t), y(t), z(t) \rangle \in \mathbb{R}^3 be the position vector of a particle of mass mm at time tt. The gravitational force F⃗g\vec{F}_g is defined by F⃗g=mg⃗\vec{F}_g = m\vec{g}, where g⃗=⟨0,0,−g⟩\vec{g} = \langle 0, 0, -g \rangle and g=GMR2g = \frac{GM}{R^2} (assuming Earth-like gravity). By Newton's Second Law, the equation of motion is:\d2r⃗dt2=F⃗gm=g⃗\frac{d^2\vec{r}}{dt^2} = \frac{\vec{F}_g}{m} = \vec{g}. Integrating this constant acceleration yields the velocity v⃗(t)=dr⃗dt=v⃗0+g⃗t\vec{v}(t) = \frac{d\vec{r}}{dt} = \vec{v}_0 + \vec{g}t, and the position r⃗(t)=r⃗0+v⃗0t+12g⃗t2\vec{r}(t) = \vec{r}_0 + \vec{v}_0 t + \frac{1}{2}\vec{g}t^2. Specifically, for initial position r⃗0=⟨x0,y0,z0⟩\vec{r}_0 = \langle x_0, y_0, z_0 \rangle and initial velocity v⃗0=⟨v0x,v0y,v0z⟩\vec{v}_0 = \langle v_{0x}, v_{0y}, v_{0z} \rangle, the position is:\r⃗(t)=⟨x0+v0xt,y0+v0yt,z0+v0zt−12gt2⟩.\vec{r}(t) = \langle x_0 + v_{0x}t, y_0 + v_{0y}t, z_0 + v_{0z}t - \frac{1}{2}gt^2 \rangle.
Source: Wikipedia