Why Does Maximum Range Occur at 45 Degrees?

This topic is part of the HSC Physics syllabus under the section Projectile Motion.

HSC Physics Syllabus

  • Apply the modelling of projectile motion to quantitatively derive the relationships between the following variables:

– maximum height
– time of flight
– final velocity
– horizontal range of the projectile

    • Solve problems, create models and make quantitative predictions by applying the equations of motion relationships for uniformly accelerated and constant rectilinear motion

    Why 45º Give Maximum Range in Projectile Motion Explained

     

    Understanding Projectile Motion

    Projectile motion refers to the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory.

    The Basic Principles

    • Two-Dimensional Motion: Projectile motion is a form of two-dimensional motion or motion in a plane. It can be understood as two one-dimensional motions, with the horizontal motion having constant velocity and vertical motion experiencing uniform acceleration due to gravity.

    • Independence of Horizontal and Vertical Motions: The key principle in understanding projectile motion is that horizontal and vertical motions are independent of each other. That means the horizontal motion does not affect the vertical motion and vice versa.

    Why 45 Degrees is Optimal for Maximum Range

    When a projectile is launched, its velocity can be broken down into two components: horizontal (`v_x`) and vertical (`v_y`). The angle of launch affects these components.
     
    The range of a projectile is given by the formula (derivation explained in the video):
    $$R = \frac{v^2 \sin{2\theta}}{g}$$
    where
    • `R` is range
    • `v` is the initial velocity
    • `\theta` is the launch angle
    • `g` is the acceleration due to gravity.

     

    To maximise range, we need to maximise the value of `\sin2\theta`. The sine function achieves its maximum value of 1 at an angle of 90 degrees. Therefore, to maximise `\sin2\theta`,  should be 90 degrees, which makes `\theta` equal to 45 degrees.