The Optimal Launch Angle

By (2.1.1), the Range depends on the Launch angle only through $\sin(2\explain{motion.launch-angle}{\theta})$. The previous chapter derives that equation. For $0^{\circ} \le \explain{motion.launch-angle}{\theta} \le 90^{\circ}$, the argument $2\explain{motion.launch-angle}{\theta}$ is between $0^{\circ}$ and $180^{\circ}$. On this interval, the sine has its maximum value, $1$, at $90^{\circ}$. So for a fixed launch speed, the range is largest at $2\explain{motion.launch-angle}{\theta} = 90^{\circ}$, i.e., at $\explain{motion.launch-angle}{\theta} = 45^{\circ}$. The explorer in Figure 2.1.1 shows this result. Diagram 2.2.1 shows the inputs to the search for this launch angle.

A polynomial maximum, no calculus

This section finds the same launch angle as the maximum of a polynomial. For $0^{\circ} \le \explain{motion.launch-angle}{\theta} \le 90^{\circ}$, the product $\sin(\explain{motion.launch-angle}{\theta})\cos(\explain{motion.launch-angle}{\theta})$ is not negative. So the range is largest where the square of this product is largest. Let $\explain{x}{u}$ be the Squared sine of the launch angle: $\explain{x}{u} = \sin^{2}(\explain{motion.launch-angle}{\theta})$. Rewriting the square of the product in terms of $\explain{x}{u}$ gives a second-order polynomial:

$$\begin{aligned} \sin^{2}(\explain{motion.launch-angle}{\theta})\,\cos^{2}(\explain{motion.launch-angle}{\theta}) &= \sin^{2}(\explain{motion.launch-angle}{\theta})\left(1 - \sin^{2}\left(\explain{motion.launch-angle}{\theta}\right)\right) \quad\text{Pythagorean identity}\\ &= \explain{x}{u}\left(1 - \explain{x}{u}\right) \quad\text{definition of }\explain{x}{u}\\ &= \explain{x}{u} - \explain{x}{u}^{2} \end{aligned}$$

(2.2.1)

The graph of this polynomial is a downward parabola with roots at $\explain{x}{u} = 0$ and $\explain{x}{u} = 1$. A downward parabola has its maximum halfway between its roots, so the polynomial is largest at $\explain{x}{u} = \tfrac{1}{2}$. Converting this value of $\explain{x}{u}$ back to a launch angle gives the Optimal launch angle:

$$\begin{aligned} \explain{theta-star}{\theta^{*}} &= \arcsin\left(\sqrt{\explain{x}{u}}\right) \qquad\text{definition of }\explain{x}{u}\text{, with }\sin(\explain{motion.launch-angle}{\theta}) \ge 0\\ &= \arcsin\left(\tfrac{1}{\sqrt{2}}\right) \qquad\text{vertex at }\explain{x}{u} = \tfrac{1}{2}\\ &= 45^{\circ} \end{aligned}$$

(2.2.2)

Setting the derivative of the polynomial to zero gives the same vertex (show how). But the symmetry of the parabola gives the vertex without a derivative. In (2.1.1), the launch speed appears only in the factor $\explain{motion.launch-speed}{v_0}^{2}$, which does not depend on the launch angle. A positive constant factor does not change the position of the maximum. So the Optimal launch angle is $45^{\circ}$ for every launch speed. ^a checks the vertex numerically and graphically.

Complementary launch angles give equal ranges

Two launch angles are complementary if their sum is $90^{\circ}$. The Complementary launch angle of $\explain{motion.launch-angle}{\theta}$ is $\explain{theta-complement}{\theta'} = 90^{\circ} - \explain{motion.launch-angle}{\theta}$. On this page the prime marks this angle and does not mean a derivative. For complementary launch angles, $\sin(2\explain{theta-complement}{\theta'}) = \sin\left(180^{\circ} - 2\explain{motion.launch-angle}{\theta}\right) = \sin(2\explain{motion.launch-angle}{\theta})$. So two complementary launch angles give equal ranges. For example, the range at a launch angle of $30^{\circ}$ equals the range at a launch angle of $60^{\circ}$. The Optimal launch angle is the only launch angle that is its own complement.

Diagram 2.2.1 Inputs to the launch-angle search

Diagram 2.2.1Inputs to the launch-angle search. The launch speed stays fixed while each launch angle is tried in turn, and the ranges they give are compared.

Figure 2.2.1Range vs. launch angle explorer. The range at every launch angle for one launch speed; the dashed line marks 45°, where the range is largest.

Controls of Figure 2.2.1

Move the launch speed slider to redraw the curve, and the launch angle slider to move the dot along it.

Launch speed (m/s)Exact launch speed in meters per second

Launch angle (degrees)Exact launch angle in degrees

The curve is symmetric about $45^{\circ}$. A launch angle of $20^{\circ}$ and a launch angle of $70^{\circ}$ give the same range. The trajectory at $20^{\circ}$ is lower and has a shorter time of flight than the trajectory at $70^{\circ}$.

Checking the polynomial vertex: its values on both sides of the vertex, and its downward-parabola shape

Evaluating the polynomial at the vertex, $\explain{x}{u} = \tfrac{1}{2}$, gives its maximum value:

$$\explain{x}{u}(1-\explain{x}{u})\big|_{\explain{x}{u}=1/2}=\tfrac{1}{4}$$

(2.2.4)

Two values of $\explain{x}{u}$ at equal distances from the vertex give equal values of the polynomial, and both values are smaller than the maximum.

Table 2.2.1Polynomial values around the vertex. The polynomial at five inputs placed symmetrically about the vertex, 1/2: inputs equally far from it give equal values.

Squared sine of the launch angle	Value of the polynomial
$0$	$0$
$\tfrac{1}{4}$	$\tfrac{3}{16}$
$\tfrac{1}{2}$	$\tfrac{1}{4}$
$\tfrac{3}{4}$	$\tfrac{3}{16}$
$1$	$0$

Figure 2.2.2The polynomial is a downward parabola. From 0 at input 0 it rises to its maximum, 0.25, at input 0.5, and falls symmetrically back to 0 at input 1.

(2.2.4), Table 2.2.1, and Figure 2.2.2 are inside the memo. A reference to each of them also works while the memo is closed.

Knowledge check 2.2.1 Optimal launch angle

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