Projectile Motion as Projection from Ground

Projectile motion is a form of motion experienced by an object that is thrown near the Earth's surface and moves along a curved path under the action of gravity only. When a projectile is launched from the ground, it follows a parabolic trajectory until it returns to the ground. This type of motion is two-dimensional and can be analyzed by considering the horizontal and vertical components of the motion separately.

Key Concepts in Projectile Motion

Projectile motion can be broken down into two components: horizontal motion and vertical motion. The horizontal motion is uniform, meaning the horizontal velocity remains constant throughout the flight of the projectile. The vertical motion is uniformly accelerated, with the acceleration due to gravity acting downwards.

Horizontal Motion

• Velocity: Constant, as there are no horizontal forces acting (assuming air resistance is negligible).
• Displacement: Linear with time.

Vertical Motion

• Velocity: Changes due to gravity.
• Displacement: Follows a parabolic path.

Important Formulas

The motion of a projectile can be described by the following equations:

• Horizontal displacement: $x = v_{0x} t$
• Vertical displacement: $y = v_{0y} t - \frac{1}{2} g t^2$
• Horizontal velocity: $v_{x} = v_{0x}$
• Vertical velocity: $v_{y} = v_{0y} - g t$
• Range of the projectile: $R = \frac{v_{0}^2 \sin(2\theta)}{g}$
• Maximum height: $H = \frac{v_{0}^2 \sin^2(\theta)}{2g}$
• Time of flight: $T = \frac{2 v_{0} \sin(\theta)}{g}$

Where:

• $v_{0x}$ and $v_{0y}$ are the initial horizontal and vertical velocities, respectively.
• $v_{0}$ is the initial velocity of the projectile.
• $\theta$ is the angle of projection with the horizontal.
• $g$ is the acceleration due to gravity (approximately $9.81 \, \text{m/s}^2$).
• $t$ is the time.

Table of Differences and Important Points

Aspect	Horizontal Motion	Vertical Motion
Initial Velocity	$v_{0x} = v_{0} \cos(\theta)$	$v_{0y} = v_{0} \sin(\theta)$
Acceleration	$0 \, \text{m/s}^2$ (no acceleration)	$g \, \text{m/s}^2$ (downwards)
Velocity	Constant	Changes with time
Displacement	Linear with time	Parabolic path
Time of Flight	Same for both components	Same for both components
Range	Depends on $v_{0x}$ and time	Not applicable
Maximum Height	Not applicable	Depends on $v_{0y}$

Examples

Example 1: Calculating Range and Maximum Height

A projectile is launched from the ground with an initial velocity of $20 \, \text{m/s}$ at an angle of $45^\circ$ with the horizontal. Calculate the range and maximum height of the projectile.

Solution:

1. Calculate the initial horizontal and vertical velocities:

   • $v_{0x} = v_{0} \cos(\theta) = 20 \cos(45^\circ) = 14.14 \, \text{m/s}$
   • $v_{0y} = v_{0} \sin(\theta) = 20 \sin(45^\circ) = 14.14 \, \text{m/s}$

2. Calculate the range of the projectile:

   • $R = \frac{v_{0}^2 \sin(2\theta)}{g} = \frac{20^2 \sin(90^\circ)}{9.81} \approx 40.82 \, \text{m}$

3. Calculate the maximum height of the projectile:

   • $H = \frac{v_{0}^2 \sin^2(\theta)}{2g} = \frac{20^2 \sin^2(45^\circ)}{2 \cdot 9.81} \approx 10.20 \, \text{m}$

Example 2: Calculating Time of Flight

Using the same initial conditions as Example 1, calculate the time of flight of the projectile.

Solution:

• $T = \frac{2 v_{0} \sin(\theta)}{g} = \frac{2 \cdot 20 \sin(45^\circ)}{9.81} \approx 2.04 \, \text{s}$

The projectile will be in the air for approximately 2.04 seconds before it hits the ground.

Understanding projectile motion as projection from the ground is essential for solving problems related to objects launched at an angle. By breaking the motion into horizontal and vertical components, we can use the kinematic equations to predict the projectile's behavior throughout its flight.