P02-060 – Range of the Projectile
Typical question — Finding the range $R$
Kinematic equations for velocity:
\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:
\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}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 $x_0=y_0=0$.
We label $t_2$the time at which the projectile reaches the ground which is where $y\left(t\right)=0$. Solving for the range can be done according to other following method.
Method 1: using $x\left(t\right)$ and $y\left(t\right)$
We find the range by solving $y\left(t_2\right)=0$ to get $t_2$ and the substitute $t_2$ back into the equation $x\left(t\right)$.
\begin{aligned}
y\left(t_2\right)=0\ \ \ \ &\Leftrightarrow \ \ \ -\frac{1}{2}gt^2_2+v_0{\mathrm{sin} \left(\theta \right)\ }t_2=0 \\
\\
&\Leftrightarrow \ \ \ \ \ t_2\left(-\frac{g}{2}t_2+v_0{\mathrm{sin} \left(\theta \right)\ }\right)=0 \\
\\
&\Leftrightarrow \ \ \ \ \ t_2=0\ \ \ \ \ or\ \ \ \ \ t_2=\frac{2v_0{\mathrm{sin} \left(\theta \right)\ }}{g}
\end{aligned}By ruling out $t_2=0$ as a solution (since represents the initial position of the projectile), we derive that the range is equal to
\begin{aligned}
R&=x\left(t_2\right) \
&=v_0{\mathrm{cos} \left(\theta \right)\ }\left(\frac{2v_0{\mathrm{sin} \left(\theta \right)\ }}{g}\right) \
&=\frac{2v^2_0{\mathrm{sin} \left(\theta \right)\ }{\mathrm{cos} \left(\theta \right)\ }}{g}
\end{aligned}The range of the projectile is therefore given by
\boxed{R=\frac{2v^2_0{\mathrm{sin} \left(\theta \right)\ }{\mathrm{cos} \left(\theta \right)\ }}{g}}Note: you will usually find the range written in its most condensed form which is
\boxed{R=\frac{v^2_0{\mathrm{sin} \left(2\theta \right)\ }}{g}}where we used the trig identity ${\mathrm{sin} \left(2\theta \right)\ }=2{\mathrm{sin} \left(\theta \right)\ }{\mathrm{cos} \left(\theta \right)\ }$.
Note: you might notice that the time $t_2$ to reach the ground is exactly double the time $t_1$ to reach maximum height. This is ONLY TRUE because we are considering a flat terrain which makes the parabolic path symmetric. If the projectile were to be launched off a cliff for example, this would no longer be the case.
Method 2: using $y\left(x\right)$ directly
If you have derived the equation of the path $y\left(x\right)$, a faster alternative is to simply solve $y\left(R\right)=0$ as it gives you the range $R$ directly.
\begin{aligned}
y\left(R\right)=0\ \ \ \ &\Leftrightarrow \ \ \ \ \ -\frac{g}{2v^2_0{{\mathrm{cos}}^2 \left(\theta \right)\ }}R^2+{\mathrm{tan} \left(\theta \right)\ }R=0 \\
\\
&\Leftrightarrow \ \ \ \ \ R\left(-\frac{g}{2v^2_0{{\mathrm{cos}}^2 \left(\theta \right)\ }}R+{\mathrm{tan} \left(\theta \right)\ }\right)=0 \\
\\
&\Leftrightarrow \ \ \ \ \ R=0\ \ \ \ \ or\ \ \ \ \ R=\frac{2v^2_0{{\mathrm{cos}}^2 \left(\theta \right)\ }{\mathrm{tan} \left(\theta \right)\ }}{g}
\end{aligned}Ruling out $R=0$ for obvious reasons, we conclude that the range is equal to
\boxed{R=\frac{2v^2_0{{\mathrm{cos}}^2 \left(\theta \right)\ }{\mathrm{tan} \left(\theta \right)\ }}{g}=\frac{2v^2_0{\mathrm{sin} \left(\theta \right)\ }{\mathrm{cos} \left(\theta \right)\ }}{g}}