# Characteristics of Projectile Motion Projectile motion refers to the curved path an object follows when launched near Earth's surface. The motion can be described by three key characteristics: the maximum height reached, the total time in the air, and the horizontal distance covered. By analyzing the vertical and horizontal components separately, we can derive simple formulas for each quantity. The following derivations assume ideal conditions where air resistance is negligible and the acceleration due to gravity ($g$) is constant. --- ## 1. Maximum Height of a Projectile (H) The maximum height is the greatest vertical distance reached by the projectile from its launch point. **Key Principle**: At the peak of its trajectory, the vertical component of the projectile's velocity is momentarily zero ($v_{fy} = 0$). We derive the formula using the third equation of motion for the vertical journey: $$2as = v_f^2 - v_i^2$$ **Initial vertical velocity**: $v_{iy} = V_i \sin\theta$ **Final vertical velocity (at peak)**: $v_{fy} = 0$ **Acceleration**: $a = -g$ **Displacement**: $s = H$ Substituting these values: $$2(-g)H = (0)^2 - (V_i \sin\theta)^2$$ $$-2gH = -V_i^2 \sin^2\theta$$ Solving for $H$: $$H = \frac{V_i^2 \sin^2\theta}{2g}$$ --- ## 2. Time of Flight (T) The time of flight is the total duration the projectile remains in the air from launch to impact. **Key Principle**: For a projectile landing at the same vertical level it was launched from, the total vertical displacement is zero ($S_y = 0$). We use the second equation of motion for the entire vertical journey: $$S = v_i t + \frac{1}{2} a t^2$$ **Vertical displacement**: $S_y = 0$ **Initial vertical velocity**: $v_{iy} = V_i \sin\theta$ **Acceleration**: $a = -g$ **Time**: $t = T$ Substituting the values: $$0 = (V_i \sin\theta)T + \frac{1}{2}(-g)T^2$$ $$\frac{1}{2}gT^2 = (V_i \sin\theta)T$$ Since $T \neq 0$, we divide both sides by $T$: $$\frac{1}{2}gT = V_i \sin\theta$$ Solving for $T$: $$T = \frac{2V_i \sin\theta}{g}$$ This is also twice the time taken to reach the maximum height. --- ## 3. Horizontal Range of a Projectile (R) The horizontal range is the total horizontal distance covered by the projectile. **Key Principle**: The horizontal component of velocity ($v_x$) is constant throughout the flight because there is no horizontal acceleration ($a_x = 0$). The horizontal distance is velocity multiplied by time: $$R = v_x \times T$$ **Horizontal velocity**: $v_x = V_i \cos\theta$ **Time of flight**: $T = \frac{2V_i \sin\theta}{g}$ Substituting: $$R = (V_i \cos\theta) \left( \frac{2V_i \sin\theta}{g} \right)$$ $$R = \frac{V_i^2 (2 \sin\theta \cos\theta)}{g}$$ Using the trigonometric identity $\sin(2\theta) = 2\sin\theta\cos\theta$: $$R = \frac{V_i^2 \sin(2\theta)}{g}$$ --- ## Conditions for Maximum Range For a fixed initial speed ($V_i$), the range $R$ depends on the launch angle $\theta$. The range is maximized when $\sin(2\theta)$ is at its maximum value, which is 1. $$\sin(2\theta) = 1$$ $$2\theta = \sin^{-1}(1) = 90^\circ$$ $$\theta = 45^\circ$$ Therefore, the maximum range is achieved at a launch angle of $45^\circ$. The formula for maximum range is: $$R_{max} = \frac{V_i^2}{g}$$ ### Same Range for Complementary Angles The range formula shows that the same range can be achieved for two different launch angles that are complementary (add up to $90^\circ$). This is because $\sin(2\theta) = \sin(180^\circ - 2\theta) = \sin(2(90^\circ - \theta))$. For example: - An angle of $30^\circ$ and an angle of $60^\circ$ will produce the same horizontal range. - An angle of $15^\circ$ and an angle of $75^\circ$ will also produce the same horizontal range. The higher angle will result in a much higher trajectory and longer time of flight, while the lower angle produces a flatter trajectory with shorter time of flight. --- | **Characteristic** | **Formula** | **Condition for Maximum** | | :--- | :--- | :--- | | **Maximum Height (H)** | $H = \frac{V_i^2 \sin^2\theta}{2g}$ | $\theta = 90^\circ$ (launched straight up) | | **Time of Flight (T)** | $T = \frac{2V_i \sin\theta}{g}$ | $\theta = 90^\circ$ | | **Horizontal Range (R)** | $R = \frac{V_i^2 \sin(2\theta)}{g}$ | $\theta = 45^\circ$ | ---