Introduction to Projectile Motion
In physics, a projectile is any object thrown into the air that moves under the influence of gravity alone. The motion of a projectile is governed by the equations of motion for constant acceleration.
The Horizontal Range Formula
The horizontal range $R$ of a projectile launched with an initial velocity $v$ at an angle $\theta$ with the horizontal is given by the formula:
$$R = \frac{v^2 \sin(2\theta)}{g}$$
where $g$ is the acceleration due to gravity.
Why Do Complementary Angles Produce the Same Range?
Consider two angles, $\theta_1$ and $\theta_2$, such that $\theta_1 + \theta_2 = 90^\circ$. We can write $\theta_2 = 90^\circ - \theta_1$.
If we substitute $\theta_2$ into the range formula:
$$R_2 = \frac{v^2 \sin(2(90^\circ - \theta_1))}{g}$$ $$R_2 = \frac{v^2 \sin(180^\circ - 2\theta_1)}{g}$$
Using the trigonometric identity $\sin(180^\circ - x) = \sin(x)$, we get:
$$R_2 = \frac{v^2 \sin(2\theta_1)}{g} = R_1$$
This proves that any two angles that add up to $90^\circ$ (complementary angles) result in the exact same horizontal range for a given initial velocity.
Solving the Problem
Given the projection angle $\theta_1 = 18^\circ$, we look for another angle $\theta_2$ such that:
$$\theta_1 + \theta_2 = 90^\circ$$ $$18^\circ + \theta_2 = 90^\circ$$ $$\theta_2 = 90^\circ - 18^\circ = 72^\circ$$
Thus, a projectile fired at $72^\circ$ will cover the same horizontal range as one fired at $18^\circ$, assuming the initial speed is identical.