[b][size=150]Position, velocity and acceleration[/size][/b][br][br]Suppose an object is moving in 3D space. We can describe its motion by specifying the position of the object at time [math]t[/math]. It can be expressed as a vector-valued function [math]\vec{r}(t)=\langle x(t),y(t),z(t)\rangle[/math] (the position vector of the object) for time [math]t \geq 0[/math]. [br][br]The parametric curve defined by [math]\vec{r}(t)[/math] is the [b]trajectory[/b] of the moving object.[br][br]The [b]velocity[/b] of the object at time [math]t[/math] is [math]\vec{v}(t)=\vec{r'}(t)=\langle x'(t),y'(t),z'(t)\rangle[/math]. It measures the infinitesimal change of position of the object at time [math]t[/math]. The [b]speed[/b] is the norm of velocity i.e. [math]|\vec{v}(t)|[/math].[br][br]The [b]acceleration[/b] of the object at time [math]t[/math] is [math]\vec{a}(t)=\vec{v'}(t)=\vec{r''}(t)=\langle x''(t),y''(t),z''(t)\rangle[/math], which is the infinitesimal change of velocity of the object at time [math]t[/math].[br][br][br][br][br]

[b]A Projectile motion in 3D space[/b][br][br]Suppose an object is launched from [math](0,0,0)[/math] at the launch angle [math]\alpha[/math], where [math]0< \alpha < 90\degree[/math] in the direction [math](\cos\theta, \sin\theta)[/math] on the xy-plane (as shown in the applet below). Let [math]s[/math] be the initial speed of launching the object. We have[br][br][math]\vec{r}(0)=\langle x(0),y(0),z(0)\rangle=\langle 0,0,0\rangle[/math][br][math]\vec{v}(0)=\langle x'(0),y'(0),z'(0)\rangle=\langle s\cos\alpha\cos\theta, s\cos\alpha\sin\theta,s\sin\alpha\rangle, \ \ |\vec{v}(0)|=s[/math][br][br]We assume the object is under the downward gravitational force. Then by Newton's second law of motion, [math]\vec{a}(t)=\langle 0,0, -g\rangle[/math], where [math]g=9.8ms^{-2}[/math] is the gravitational acceleration on Earth.[br][br]Now we use integration to solve for [math]\vec{v}(t)[/math] and [math]\vec{r}(t)[/math]:[br][br][math]\vec{v}(t)=\vec{v}(0)+\int_0^t \vec{a}(t) \ dt=\langle s\cos\alpha\cos\theta, s\cos\alpha\sin\theta,s\sin\alpha\rangle+\int_0^t \langle 0,0, -g\rangle \ dt=\langle s\cos\alpha\cos\theta, s\cos\alpha\sin\theta,s\sin\alpha-gt\rangle[/math][br][br][math]\vec{r}(t)=\vec{r}(0)+\int_0^t \vec{v}(t) \ dt = \langle 0,0,0\rangle+\int_0^t \langle s\cos\alpha\cos\theta, s\cos\alpha\sin\theta,s\sin\alpha-gt\rangle \ dt = \langle st\cos\alpha\cos\theta, st\cos\alpha\sin\theta,st\sin\alpha-\frac 12 gt^2\rangle[/math][br][br]Let [math]T[/math] be the time when the object hits the ground again. Then we have[br][br][math]sT\sin\alpha-\frac 12 gT^2=0[/math][br][math]\implies T=\frac{2s\sin\alpha}g[/math][br][br]Let [math]R[/math] be the range of the projectile i.e. the distance from the origin (the launch location) to the place where the object hits the ground again. Then we have[br][br][math]R=|\vec{r}(T)|[/math][br][math]\implies R=\sqrt{ (\langle sT\cos\alpha\cos\theta)^2+(sT\cos\alpha\sin\theta)^2}=sT\cos\alpha=\frac{2s^2\sin\alpha\cos\alpha}g=\frac{s^2\sin(2\alpha)}g[/math][br][br]Let [math]H[/math] be the maximum height of the trajectory. It occurs at [math]t=\frac{T}2[/math]. Therefore, we have[br][br][math]H=s\frac{T}{2}\sin\alpha-\frac 12 g\left(\frac T2\right)^2=\frac{s^2\sin^2\alpha}g-\frac{s^2\sin^2\alpha}{2g}=\frac{s^2\sin^2\alpha}{2g}[/math][br]