Why Maximum Range Occurs at 45 Degrees?

Have you ever wondered why a projectile travels the farthest when launched at a 45-degree angle? Whether you're a student tackling kinematics problems or just someone intrigued by the physics of motion, this blog post is for you. We'll delve into the fundamentals of projectile motion, derive the equations that explain maximum range, and understand why that magical 45-degree angle is the key to achieving it.

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Table of Contents

• Introduction to Projectile Motion
• Understanding the Components of Motion
• Deriving the Range Equation
• Maximizing the Range: Why 45 Degrees?
• Practical Considerations
• Common Misconceptions
• Conclusion

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Introduction to Projectile Motion

Projectile motion is a form of motion where an object moves in a bilaterally symmetrical, parabolic path under the action of gravity alone. Classic examples include a football being kicked, a cannonball fired from a cannon, or water sprayed from a hose.

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Understanding the Components of Motion

In projectile motion, the object's motion can be broken down into two independent components:

1. Horizontal Motion ($x$-axis):
   • Velocity ($v_x$) remains constant (assuming no air resistance).
   • Acceleration ($a_x$) is zero.
2. Vertical Motion ($y$-axis):
   • Velocity ($v_y$) changes due to gravity.
   • Acceleration ($a_y$) is $-g$, where $g=9.8{\text{m/s}}^2$.

Key Equations:

$$\text{Horizontal Displacement:}x=v_{0x}t$$

$$\text{Vertical Displacement:}y=v_{0y}t-\frac{1}{2}g{t}^2$$

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Deriving the Range Equation

Objective: Find the horizontal distance ($R$) the projectile travels before landing back on the same vertical level.

Step 1: Break Down the Initial Velocity

Let $v_0$ be the initial launch speed and $\theta$ be the launch angle.

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

Step 2: Calculate Time of Flight ($t$)

The time the projectile is in the air depends on the vertical motion.

At landing, the vertical displacement $y=0$.

Using the vertical displacement equation:

$$0=v_{0y}t-\frac{1}{2}g{t}^2$$

Solve for $t$:

$$t=\frac{2v_{0y}}{g}=\frac{2v_0\sin \theta }{g}$$

Step 3: Determine the Range ($R$)

Using the horizontal displacement equation:

$$R=v_{0x}t$$

Substitute $v_{0x}$ and $t$:

$$R=(v_0\cos \theta )\left(\frac{2v_0\sin \theta }{g}\right)$$

Simplify:

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

Using the trigonometric identity $\sin 2\theta =2\sin \theta \cos \theta$:

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

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Maximizing the Range: Why 45 Degrees?

To find the angle $\theta$ that maximizes $R$, consider $\sin 2\theta$ in the range equation:

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

• Maximum Value of $\sin 2\theta$: The sine function reaches its maximum value of 1 when $2\theta ={90}^{\circ }$.

Therefore:

$$2\theta ={90}^{\circ }⟹\theta ={45}^{\circ }$$

Conclusion: The projectile achieves maximum range when launched at a 45-degree angle.

Remember:

• Break Down Vectors: Always resolve initial velocity into $v_{0x}$ and $v_{0y}$.
• Use Key Equations: Familiarize yourself with the range equation and how to derive it.
• Consider Real-World Factors: Air resistance and varying initial heights can alter the optimal angle.

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Keep exploring and questioning! Physics is not just about memorizing equations but understanding the principles that govern the natural world. If you have any questions or thoughts, feel free to leave them in the comments below.

Happy Studying!