Understanding Projectile Range and Angle of Projection

When an object is launched into the air, like a ball thrown or a cannonball fired, it follows a curved path called a trajectory. This object is known as a projectile. The horizontal distance it travels from the launch point to where it lands is called the range of the projectile.

The range of a projectile depends on several factors, including the initial speed of the projectile, the angle at which it is launched (angle of projection), and the acceleration due to gravity. For a given initial speed and ignoring air resistance, the range is determined by the angle of projection.

Formula for Projectile Range

The formula for the horizontal range ($R$) of a projectile launched with an initial velocity ($v_0$) at an angle ($\theta$) with respect to the horizontal is given by:

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

where $v_0$ is the initial velocity, $\theta$ is the angle of projection, and $g$ is the acceleration due to gravity.

Maximizing Projectile Range

To find the angle of projection ($\theta$) for which the range ($R$) is maximum, we need to look at the formula. Assuming the initial velocity ($v_0$) and the acceleration due to gravity ($g$) are constant, the range ($R$) is directly proportional to $\sin(2\theta)$.

For the range $R$ to be maximum, the value of $\sin(2\theta)$ must be maximum. The maximum value that the sine function can take is 1. This occurs when the angle is $90^\circ$, $90^\circ + 360^\circ$, $-270^\circ$, etc. We are interested in the angle $2\theta$ for the sine function.

So, we need $\sin(2\theta) = 1$.

This happens when:

$$2\theta = 90^\circ$$

Now, we can find the angle of projection ($\theta$) by dividing by 2:

$$\theta = \frac{90^\circ}{2}$$

$$\theta = 45^\circ$$

Therefore, the range of a projectile is maximum when the angle of projection is $45^\circ$.

Analyzing the Given Options for Maximum Range

Let's evaluate the range for the angles given in the options, keeping in mind that maximum range occurs at $45^\circ$.

• Option 1: $60^\circ$

  At $\theta = 60^\circ$, $2\theta = 120^\circ$.

  $\sin(120^\circ) = \sin(180^\circ - 60^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2} \approx 0.866$

  Range $R \propto 0.866$

• Option 2: $45^\circ$

  At $\theta = 45^\circ$, $2\theta = 90^\circ$.

  $\sin(90^\circ) = 1$

  Range $R \propto 1$ (Maximum value)

• Option 3: $90^\circ$

  At $\theta = 90^\circ$, $2\theta = 180^\circ$.

  $\sin(180^\circ) = 0$

  Range $R \propto 0$. When launched vertically ($90^\circ$), the projectile goes up and comes straight down, so the horizontal range is zero.

• Option 4: $30^\circ$

  At $\theta = 30^\circ$, $2\theta = 60^\circ$.

  $\sin(60^\circ) = \frac{\sqrt{3}}{2} \approx 0.866$

  Range $R \propto 0.866$

Comparing the $\sin(2\theta)$ values:

• $60^\circ \implies \sin(120^\circ) \approx 0.866$
• $45^\circ \implies \sin(90^\circ) = 1$
• $90^\circ \implies \sin(180^\circ) = 0$
• $30^\circ \implies \sin(60^\circ) \approx 0.866$

The maximum value of $\sin(2\theta)$ is 1, which occurs at $\theta = 45^\circ$. Notice that $30^\circ$ and $60^\circ$ give the same range because $\sin(2 \times 30^\circ) = \sin(60^\circ)$ and $\sin(2 \times 60^\circ) = \sin(120^\circ) = \sin(180^\circ - 60^\circ) = \sin(60^\circ)$. Angles complementary to $45^\circ$ (i.e., angles $\theta$ and $90^\circ - \theta$) give the same range, except for $45^\circ$ itself, which is complementary to itself and gives the maximum range.

Based on this analysis, the angle of projection that results in the maximum range is $45^\circ$.

Revision Table: Projectile Motion Formulas

Additional Information on Projectile Motion

Understanding projectile motion involves more than just the range. Other key aspects include the time the projectile stays in the air (time of flight) and the highest point it reaches (maximum height).

• Time of Flight: The time of flight ($T$) is the total time from launch until the projectile hits the ground. It depends on the vertical component of the initial velocity and gravity. The formula is $T = \frac{2 v_0 \sin(\theta)}{g}$. The maximum time of flight occurs when $\sin(\theta)$ is maximum, which is at $\theta = 90^\circ$ (vertical launch).
• Maximum Height: The maximum height ($H$) reached by the projectile depends only on the vertical component of the initial velocity. The formula is $H = \frac{v_0^2 \sin^2(\theta)}{2g}$. Maximum height is achieved when $\sin(\theta)$ is maximum, which is also at $\theta = 90^\circ$.
• Symmetry: The trajectory of a projectile launched on level ground is a parabola (ignoring air resistance). The path is symmetrical about the highest point. The time taken to reach the maximum height is half the total time of flight.
• Effect of Angle:
  • Small angles (e.g., $15^\circ$, $30^\circ$) result in shorter range and lower height.
  • The range is the same for angles $\theta$ and $(90^\circ - \theta)$, provided the initial speed is the same. For example, the range at $30^\circ$ is the same as the range at $60^\circ$.
  • The maximum range occurs uniquely at $45^\circ$.
  • A vertical launch ($90^\circ$) gives zero range and maximum height/time of flight.
  • A horizontal launch ($0^\circ$) from a height gives range but no time of flight as calculated by the standard formula (this formula is for launch from ground to ground).