U3AOS1 Topic 4: Projectile Motion

$x_{horizontal}=ut\ast cos\left(\theta \right)$

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$x_{horizontal} = ut*cos(\theta)$

Introduction

Projectile motion describes the trajectory of an object that is launched into the air and moves under the influence of gravity and its initial velocity. This type of motion is crucial in fields such as physics, engineering, and sports, and it helps us understand how objects like thrown balls, launched missiles, and even space probes travel through the air.

Basic Concepts

Projectile motion can be decomposed into two independent components: horizontal and vertical motion. Understanding these components separately simplifies the analysis.

1. Horizontal Motion:

   • Constant Velocity: In the absence of air resistance, the horizontal component of velocity remains constant throughout the motion. This is because no horizontal force acts on the projectile (assuming air resistance is negligible). Thus, the horizontal distance traveled (range) can be calculated as $x = v_x t$, where $v_x$ is the horizontal component of the initial velocity and $t$ is the time of flight.

2. Vertical Motion:

   • Uniform Acceleration: The vertical motion is influenced by gravity, which provides a constant acceleration downward. This acceleration, denoted as $g$, has a value of approximately $9.8\ \text{m/s}^2$ on Earth. The vertical displacement can be calculated using the kinematic equations of motion, such as $y = v_y t - \frac{1}{2} g t^2$, where $v_y$ is the vertical component of the initial velocity.

Key Equations and Parameters

1. Initial Velocity Components:

   • If an object is launched with an initial velocity $v_0$ at an angle $\theta$ from the horizontal, the initial velocity components are: $v_x = v_0 \cos \theta$ $v_y = v_0 \sin \theta$

2. Time of Flight (T):

   • The time the projectile remains in the air depends on the vertical motion. For a projectile launched and landing at the same height, the time of flight can be found using: $T = \frac{2 v_y}{g} = \frac{2 v_0 \sin \theta}{g}$

3. Maximum Height (H):

   • The maximum height is reached when the vertical velocity becomes zero. The height can be calculated using: $H = \frac{v_y^2}{2g} = \frac{(v_0 \sin \theta)^2}{2g}$

4. Range (R):

   • The horizontal distance traveled by the projectile is given by: $R = v_x T = \frac{v_0^2 \sin 2\theta}{g}$

Trajectory

The trajectory of a projectile is a parabola. This shape results from the combination of constant horizontal velocity and the uniformly accelerated vertical motion. The path can be described by combining the horizontal and vertical motions into a single equation, which generally takes the form: $y = x \tan \theta - \frac{g x^2}{2 (v_0 \cos \theta)^2}$

Applications

Projectile motion principles are applied in various domains:

• Sports: In basketball, understanding the projectile path helps in shooting accurately. Athletes use these principles to determine the optimal angle and velocity for scoring.
• Engineering: Engineers use projectile motion concepts in designing trajectories for missiles and projectiles to ensure they hit targets accurately.
• Space Exploration: Calculating the trajectories of spacecraft involves understanding and applying projectile motion principles, often under the influence of different gravitational fields.

There are two main methods for solving projectile questions:

SUVAT Method

Angle Method

$v = u + at$ $$ \[ v= u+ at \] \[ s = ut+ \frac{1}{2}at^2\] \[s = vt - \frac{1}{2}at^2\] \[s = \frac{t}{2}(v+u)\] \[ v^2 = u^2+2as\]

Key Formula: \[ h_{\text{max}}=\frac{1}{2g}(u \times \sin(\theta))^2\] \[ \text{range} = \frac{u^2}{g} \sin(2 \theta)\] \[ \text{t at} h_{\text{max}}= \frac{u}{g}\sin(\theta)\] $h_{max} = \frac{1}{2g}(u*sin(\theta))^2$ $v$ $\backslash (\; \backslash text\{velocity\; at\; t:\}\backslash )$ \[ v_{\text{vertical}}= u \times \sin(\theta) - gt\] \[ v_{\text{horizontal}}= u \times \cos(\theta)\] $v_{vertical} = u*sin(\theta) - gt$ \( \text{distance at t:}\) \[ x_{\text{vertical}} = ut \times \sin(\theta) - \frac{1}{2}gt^2\] \[ x_{\text{horizontal}} = ut \cos(\theta)\] $x_{vertical} = ut*sin(\theta) - \frac{1}{2}gt^2$             $h_{max}=\frac{1}{2g}(u\ast sin(\theta ){)}^2$ ${�}_{ℎ���������}=�\ast ���(�)<span\; style="font-family:\; KaTeX\_Math;\; font-style:\; italic;"><br></span>$

    

Although there are more formulas for the angles method, it is an easier method to use given the initial speed and launch angle are provided.

However, they may not be used in all scenarios therefore it is important to know both methods of solving.

Solving with SUVAT requires breaking down equations into horizontal and vertical components.

IMPORTANT NOTE

Air resistance is generally neglected.

Hence, the horizontal speed will be the same throughout the journey and only the vertical speed will change.

However, if there is air resistance the object will not travel as far or as high.

Use the simulation below to check your calculations and practice