The range of a projectile is the horizontal distance it travels while in motion. For a projectile launched with an initial velocity $v_0$ at an angle $\theta$ with respect to the horizontal, the range $R$ is given by the formula:

$R = \frac{v_0^2 \sin(2\theta)}{g}$

where:

• $v_0$ is the initial velocity of the projectile,
• $\theta$ is the angle of launch,
• $g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$ on Earth).

Derivation of the Formula

1. Horizontal and Vertical Components of the Initial Velocity:

   • Horizontal component: $v_{0x} = v_0 \cos \theta$
   • Vertical component: $v_{0y} = v_0 \sin \theta$

2. Time of Flight:

   • The time $t$ the projectile stays in the air is determined by the vertical motion.
   • Using the equation of motion $y = v_{0y} t - \frac{1}{2} g t^2$, and setting $y = 0$ for the range (assuming it lands at the same height from which it was launched), we get: $t = \frac{2 v_{0y}}{g} = \frac{2 v_0 \sin \theta}{g}$

3. Range Calculation:

   • The horizontal distance traveled (range) $R$ is given by: $R = v_{0x} \times t$
   • Substituting $v_{0x}$ and $t$: $R = (v_0 \cos \theta) \left( \frac{2 v_0 \sin \theta}{g} \right)$
   • Simplifying the expression: $R = \frac{v_0^2 \sin 2\theta}{g}$

Would you like more details or have any questions about this?

Here are five related questions you might ask:

1. How does air resistance affect the range of a projectile?
2. What is the maximum height reached by a projectile?
3. How can you derive the time of flight for a projectile?
4. What happens to the range if the angle of projection is increased or decreased?
5. How do you calculate the initial velocity needed for a given range?

Tip: Always resolve the initial velocity into horizontal and vertical components to analyze projectile motion effectively.