Understanding Projectile Motion
In physics, a projectile is any object thrown or launched into the air that is subject only to the acceleration of gravity. When we discuss the maximum range of a projectile, we are referring to the total horizontal distance the object travels before returning to the same vertical level from which it was launched.
The Formula for Maximum Range
The range $R$ of a projectile launched with an initial velocity $v$ at an angle $\theta$ is given by the formula:
$$R = \frac{v^2 \sin(2\theta)}{g}$$
Where:
- $v$ is the initial velocity.
- $\theta$ is the angle of projection.
- $g$ is the acceleration due to gravity.
To achieve the maximum possible range, we set the projection angle to $\theta = 45^\circ$. Since $\sin(2 \times 45^\circ) = \sin(90^\circ) = 1$, the formula simplifies to:
$$R_{max} = \frac{v^2}{g}$$
Analyzing the Change
We want to know what happens to $R_{max}$ if we double the initial velocity (i.e., replacing $v$ with $2v$).
- Original Range ($R_1$): $R_1 = \frac{v^2}{g}$
- New Range ($R_2$): Substitute $2v$ into the equation: $$R_2 = \frac{(2v)^2}{g} = \frac{4v^2}{g}$$
- Comparison: We can see that $R_2 = 4 \times (\frac{v^2}{g})$, which means: $$R_2 = 4R_1$$
Conclusion
By doubling the initial velocity of a projectile, the maximum range increases by a factor of four. This quadratic relationship ($R \propto v^2$) happens because the initial velocity affects both the time the projectile stays in the air and its horizontal speed simultaneously.