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Why does a projectile launched without air resistance travel farthest at an angle of 45°?

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Diagram of a right triangle labeled with sides a, b, and hypotenuse c, alongside formulas for the six trigonometric functions-sine, cosine, tangent, cosecant, secant, and cotangent-each defined by ratios of the triangle's sides. — trigonometry Based on the definitions, various simple relationships exist among the functions. For example, csc A = 1/sin A, sec A = 1/cos A, cot A = 1/tan A, and tan A = sin A/cos A.

In the study of projectile motion, the theoretical range $R$ of an object launched in a vacuum is determined by the formula:

$R = \frac{v_{0}^{2} \sin 2 A}{g}$

In this equation, $v_{0}$ represents the initial muzzle velocity, $g$ is the acceleration due to gravity, and $A$ is the angle of elevation.1

Because the range depends on the sine of twice the angle ( $\sin 2 A$ ), the maximum value occurs when $\sin 2 A$ is at its maximum of 1. This happens when $2 A = 90^{\circ}$ , which simplifies to an angle of $A = 45^{\circ}$ . Mathematically, the function reaches this peak at $45^{\circ}$ and decreases symmetrically as the angle moves higher or lower. While this model is essential for understanding the underlying principles of physics, it assumes the absence of air resistance, which in real-world scenarios significantly alters the actual trajectory of a projectile.12