P02-070 – Launch Angle and Range Maximization

Typical question — Finding the angle that maximizes the range

The projectile achieves the greatest range when the launch angle is ${\theta }_0=45{}^\circ $.

There are two ways of going about this question and you can reason conceptually or analytically.

Method 1: conceptual reasoning

The range is maximized when the launch angle equals $45{}^\circ $ because then your initial velocity components $v_{0x}$ and $v_{0y}$ are equal.

Indeed, this is the best possible compromise between wanting to shoot very high but travel slowly to the right (big $v_{0y}$ and small $v_{0x}$ for $\theta >45{}^\circ $) and shoot very low but travel very fast to the right (small $v_{0y}$ and big $v_{0x}$ for $\theta <45{}^\circ $).

If the components of the initial velocity are equal, the projectile travels the farthest.

Method 2: analytical reasoning

Recall that the range is given by the following expression

R=\frac{v^2_0{\mathrm{sin} \left(2\theta \right)\ }}{g}

and note that we specifically choose to use the most condensed expression for $R$ as derived previously.

We can then argue that the quantity ${\mathrm{sin} \left(2\theta \right)\ }$ is at most equal to $1$ and when it is, the range is maximized. In other words,

\begin{aligned}
R=R_{max}\ \ \ \ &\Leftrightarrow \ \ \ \ \frac{v^2_0{\mathrm{sin} \left(2\theta \right)\ }}{g}=\frac{v^2_0}{g} \\
\\
&\Leftrightarrow \ \ \ \ {\mathrm{sin} \left(2\theta \right)\ }=1 \\
\\
&\Leftrightarrow \ \ \ \ 2\theta =90{}^\circ \\
\\
&\Leftrightarrow \ \ \ \ \theta =45{}^\circ
\end{aligned}

The range is therefore maximized when $\boxed{\theta =45{}^\circ }$.