Source: High school physics (Chinese)
Problem
An object is launched from level ground with a fixed initial speed $v_0$ at an angle $\theta$ above the horizontal. Air resistance is negligible.
- Prove that the horizontal range is maximum when the launch angle is $45^\circ$.
- Prove that the horizontal ranges are equal if two launch angles are complementary.
[Q1] The horizontal range $R(\theta) = \frac{v_0^2}{g}\sin(2\theta)$ is maximized when the term $\sin(2\theta)$ is maximum, which is 1. This occurs when $2\theta=90^\circ$, so the launch angle for maximum range is $\theta=45^\circ$. [Q2] For complementary angles $\theta_1$ and $\theta_2 = 90^\circ - \theta_1$, the ranges are $R_1 = \frac{v_0^2}{g}\sin(2\theta_1)$ and $R_2 = \frac{v_0^2}{g}\sin(2(90^\circ - \theta_1)) = \frac{v_0^2}{g}\sin(180^\circ - 2\theta_1)$. Since the trigonometric identity $\sin(180^\circ - \phi) = \sin(\phi)$ holds, it follows that $R_1 = R_2$.
The horizontal motion of a projectile is described by $x(t) = (v_0 \cos\theta) t$, and the vertical motion by $y(t) = (v_0 \sin\theta) t - \frac{1}{2}gt^2$. The key to solving both questions is the formula for the horizontal range $R$.
Derivation of the Horizontal Range Formula The projectile lands when its vertical position returns to zero, $y(T) = 0$, where $T$ is the total time of flight.
$$0 = (v_0 \sin\theta) T - \frac{1}{2}gT^2$$Solving for $T eq 0$ gives the time of flight:
$$T = \frac{2v_0 \sin\theta}{g}$$The horizontal range $R$ is the horizontal distance traveled during this time, $R = x(T)$.
$$R = (v_0 \cos\theta) T = (v_0 \cos\theta) \left( \frac{2v_0 \sin\theta}{g} \right) = \frac{v_0^2}{g} (2\sin\theta\cos\theta)$$Using the trigonometric double-angle identity, $2\sin\theta\cos\theta = \sin(2\theta)$, we obtain the range formula:
$$R(\theta) = \frac{v_0^2}{g} \sin(2\theta)$$[Q1] Proof of Maximum Range at $45^\circ$ The range formula is $R(\theta) = \frac{v_0^2}{g} \sin(2\theta)$. With the initial speed $v_0$ and gravity $g$ being constant, the range $R$ is a function of the launch angle $\theta$ only. The range is maximized when the term $\sin(2\theta)$ is maximum. The maximum value of the sine function is 1.
$$\sin(2\theta) = 1$$This condition is met when the argument of the sine function is $90^\circ$ (or $\pi/2$ radians).
$$2\theta = 90^\circ \implies \theta = 45^\circ$$Therefore, the horizontal range is maximum for a launch angle of $45^\circ$.
(Alternative calculus proof: The maximum is found by setting the derivative $\frac{dR}{d\theta}$ to zero: $\frac{dR}{d\theta} = \frac{2v_0^2}{g}\cos(2\theta) = 0$. This gives $\cos(2\theta)=0$, which implies $2\theta = 90^\circ$, so $\theta = 45^\circ$.)
[Q2] Proof of Equal Ranges for Complementary Angles Let two launch angles, $\theta_1$ and $\theta_2$, be complementary. This means their sum is $90^\circ$:
$$\theta_1 + \theta_2 = 90^\circ \implies \theta_2 = 90^\circ - \theta_1$$We will now compare the range for each angle using the range formula.
The range for angle $\theta_1$ is:
$$R_1 = \frac{v_0^2}{g} \sin(2\theta_1)$$The range for angle $\theta_2$ is:
$$R_2 = \frac{v_0^2}{g} \sin(2\theta_2)$$Substituting $\theta_2 = 90^\circ - \theta_1$:
$$R_2 = \frac{v_0^2}{g} \sin(2(90^\circ - \theta_1)) = \frac{v_0^2}{g} \sin(180^\circ - 2\theta_1)$$Using the trigonometric identity $\sin(180^\circ - \phi) = \sin(\phi)$, with $\phi = 2\theta_1$, we find:
$$\sin(180^\circ - 2\theta_1) = \sin(2\theta_1)$$Thus, the expression for $R_2$ simplifies to:
$$R_2 = \frac{v_0^2}{g} \sin(2\theta_1)$$By comparison, $R_1 = R_2$. This proves that the horizontal ranges are equal for complementary launch angles.