P02-020 – Equations of Motion
Equations of motion
Consider a projectile launched from the ground with an initial velocity $\overrightarrow{v_0}$ at an angle $\theta $ with respect to the ground. In this specific example, we assume that the origin of the coordinate system is placed at the launching point of the projectile. We then have $x_0=y_0=0$.
The equations of motion of the projectile refer to the kinematic equations $x\left(t\right)$ and $y\left(t\right)$ that give its position as a function of time along $x$ and $y$ respectively as well as $v_x\left(t\right)$ and $v_y\left(t\right)$ that give its velocity along $x$ and $y$ respectively.
To derive these four equations, we start with the components of acceleration and then integrate them with respect to time according to the following steps.
Step 1: integrating acceleration components to derive the components $v_x\left(t\right)$ and $v_y\left(t\right)$ of velocity
We integrate the components of acceleration with respect to time as follows
\left\{ \begin{array}{c}
a_x=0 \ \ \ \\
a_y=-g \end{array}
\right.\ \ \ \ \Rightarrow \ \ \ \ \ \left\{ \begin{array}{c}
v_x\left(t\right)=C_1 \ \ \ \ \ \ \ \ \ \ \ \ \\
v_y\left(t\right)=-gt+C_2 \end{array}
\right.where $C_1$ and $C_2$ are constant that are determined by applying the initial conditions
\left\{ \begin{array}{c}
v_x\left(0\right)=v_0{\mathrm{cos} \left(\theta \right)\ } \\
v_y\left(0\right)=v_0{\mathrm{sin} \left(\theta \right)\ } \end{array}
\right.which yields
\left\{ \begin{array}{c}
v_x\left(0\right)=v_0{\mathrm{cos} \left(\theta \right)\ } \\
v_y\left(0\right)=v_0{\mathrm{sin} \left(\theta \right)\ } \end{array}
\right.\ \ \ \ \Rightarrow \ \ \ \ \ \left\{ \begin{array}{c}
C_1=v_0{\mathrm{cos} \left(\theta \right)\ } \\
C_2=v_0{\mathrm{sin} \left(\theta \right)\ } \end{array}
\right.Thus, we conclude that the components of the velocity of the projectile are given by
\left\{ \begin{array}{c}
v_x\left(t\right)=v_0{\mathrm{cos} \left(\theta \right)\ }\ \ \ \ \ \ \ \ \ \ \ \\
v_y\left(t\right)=-gt+v_0{\mathrm{sin} \left(\theta \right)\ } \end{array}
\right.Step 2: integrating velocity components to derive the components $x\left(t\right)$ and $y\left(t\right)$ of position
We integrate the components of velocity with respect to time as follows
\left\{ \begin{array}{c}
v_x\left(t\right)=v_0{\mathrm{cos} \left(\theta \right)\ }\ \ \ \ \ \ \ \ \ \ \ \ \ \\
v_y\left(t\right)=-gt+v_0{\mathrm{sin} \left(\theta \right)\ } \end{array}
\right.\ \ \ \ \Rightarrow \ \ \ \ \ \left\{ \begin{array}{c}
x\left(t\right)=v_0{\mathrm{cos} \left(\theta \right)\ }+C_3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\
\\
\displaystyle{y\left(t\right)=-\frac{1}{2}gt^2+v_0{\mathrm{sin} \left(\theta \right)\ }t+C_4} \end{array}
\right.where $C_3$ and $C_4$ are constant that are determined by applying the initial conditions
\left\{ \begin{array}{c}
x\left(0\right)=0 \\
y\left(0\right)=0 \end{array}
\right.which yields
\left\{ \begin{array}{c}
x\left(0\right)=0 \\
y\left(0\right)=0 \end{array}
\right.\ \ \ \ \Rightarrow \ \ \ \ \ \left\{ \begin{array}{c}
x\left(t\right)=v_0{\mathrm{cos} \left(\theta \right)\ }t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\
\\
\displaystyle{y\left(t\right)=-\frac{1}{2}gt^2+v_0{\mathrm{sin} \left(\theta \right)\ }t} \end{array}
\right.Thus, we conclude that the components of the velocity of the projectile are given by
\left\{ \begin{array}{c}
x\left(0\right)=x_0 \\
y\left(0\right)=y_0 \end{array}
\right.\ \ \ \ \Rightarrow \ \ \ \ \ \left\{ \begin{array}{c}
x\left(t\right)=v_0{\mathrm{cos} \left(\theta \right)\ }t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\
\\
\displaystyle{y\left(t\right)=-\frac{1}{2}gt^2+v_0{\mathrm{sin} \left(\theta \right)\ }t} \end{array}
\right.Summary:
In general, we can repeat this process for projectile motion scenarios where $x_0$ and $y_0$ are not necessarily both zero. In which case, the kinematic equations we would derive would be
Kinematic equations for velocity:
\boxed{
\begin{aligned}
v_x\left(t\right)&=v_0{\mathrm{cos} \left(\theta \right)\ } \\
v_y\left(t\right)&=-gt+v_0{\mathrm{sin} \left(\theta \right)\ }
\end{aligned}}Kinematic equations for position:
\boxed{
\begin{aligned}
x\left(t\right)&=v_0{\mathrm{cos} \left(\theta \right)\ }t+x_0 \\
y\left(t\right)&=-\frac{1}{2}gt^2+v_0{\mathrm{sin} \left(\theta \right)\ }t+y_0
\end{aligned}}