Linear Kinematics in Two Dimensions & Projectile Motion

### Linear Kinematics in Two Dimensions & Projectile Motion - Movement in both x and y directions. - Assignment 2 practice questions. #### Learning Objectives - Projectile motion: the trajectory is a parabola. - Calculate horizontal range and its factors. - Relative height. - Equations for the range of a projectile. - Factors determining a projectile's trajectory. - Equations in x and y directions separately. - Applications and practice. #### Factors Determining Height and Distance - **Initial velocity**: Includes both magnitude (speed) and direction (angle (\theta)) at takeoff. - Angle (\theta)) controls horizontal and vertical distances. - **Projectile motion**: Airborne body subjected only to gravity and air resistance (typically air resistance is ignored). #### Trajectory - The path of a projectile is called a **parabola**. - Examples include somersaults, baseball throws, long jumps, javelin throws, and basketball shots. #### Trajectory in Outer Space (Zero Gravity) - In zero gravity, the trajectory would be a straight line. - Gravity pulls objects downward, causing the parabolic trajectory. #### Equations of Motion - **Default Assumptions**: X-direction is horizontal (positive to the right), Y-direction is vertical (positive upward). - Apply the h model in both x and y directions. ##### X-Direction Equations - Final position: x_1 = x_0 + \Delta x - Displacement: \Delta x = v_{x0} \Delta t + \frac{1}{2} a_x (\Delta t)^2 - Change of velocity: \Delta v_x = a_x \Delta t ##### Y-Direction Equations - Final position: y_1 = y_0 + \Delta y - Includes acceleration due to gravity (a_y). - Gravity is constant on Earth, acting downward. - a_y = -9.81 \frac{m}{s^2} (negative sign indicates downward direction). - Absolute value: g = \|a_g\| = 9.81 \frac{m}{s^2} #### Deriving Projectile Motion Equations (Y-Direction) - Vertical direction acceleration: a_y = -9.81 \frac{m}{s^2} - Definition of acceleration: a_y = \frac{\Delta v_y}{\Delta t} - Change of velocity: \Delta v_y = a_y \Delta t - Vertical direction displacement: \Delta y = v_{y0} \Delta t + \frac{1}{2} a_y (\Delta t)^2 #### Derivation of Vertical Displacement Equation - \Delta y is the area covered by the velocity-time curve. - Slope of the velocity curve is acceleration. - Acceleration is a negative constant. - The velocity-time curve is a straight line going down, with a negative constant slope. - The slope (tangent (\theta)) is a negative constant: tan(\theta) = -9.81 \frac{m}{s^2} #### Vertical Displacement - \Delta y is the area under the velocity-time curve. - The slope of the position-time curve is the vertical direction velocity. - Positive velocity means positive slope. - As velocity gradually decreases, the slope gradually decreases. - Calculating \Delta y (displacement): The area consists of a right triangle and a rectangle. - Final velocity v_1 = v_{y0} + \Delta v_y - Area of the right triangle: \frac{1}{2} \Delta t (-\Delta v_y) - Total displacement: \Delta y = v_{y0} \Delta t + \frac{1}{2} \Delta t (-\Delta v_y) #### Alternative Calculation of Displacement - Using the area of a trapezoid. - Area = \frac{A + B}{2} H - A = v_{y0}, B = v_{y1} = v_{y0} + \Delta v_y, H = \Delta t - \Delta y = v_{y0} \Delta t + \frac{1}{2} a_y (\Delta t)^2 #### Summary of Vertical Direction Equations - Change in velocity: \Delta v_y = a_y \Delta t , where a_y = -9.81 \frac{m}{s^2} - Vertical displacement: \Delta y = v_{y0} \Delta t + \frac{1}{2} a_y (\Delta t)^2 #### Key Points about Vertical Displacement - Acceleration due to gravity alters the projectile path only along the y-direction. - The relationship between vertical displacement and time spent ( \Delta t ) is not linear due to the squared term. #### Example - Dropping a ball. After \Delta t , displacement is D_1 . After another \Delta t , displacement is D_2 . - Due to the squared term, D_2 > 2D_1 . #### Horizontal Direction - No gravity: Acceleration in the horizontal direction is zero. - Change of velocity: \Delta v_x = a_x \Delta t = 0 - Velocity is constant: v_x = v_{x0} - Displacement increases linearly: \Delta x = v_x \Delta t #### Combining Vertical and Horizontal Motion - Instantaneous velocity can be decomposed into x and y components. - Vertical and horizontal velocities behave differently. ##### Vertical vs. Horizontal Velocity - Vertical: \Delta v_y = a_y \Delta t - Horizontal: \Delta v_x = 0 ##### Vertical vs. Horizontal Displacement - Vertical: \Delta y = v_{y0} \Delta t + \frac{1}{2} a_y (\Delta t)^2 - Horizontal: \Delta x = v_x \Delta t #### Projectile Motion Features - Horizontal velocity ( v_x ) is constant. - Vertical velocity ( v_y ) changes due to gravity. - At the apex (highest point), v_y = 0 #### Simple Cases ##### Free Fall - Only movement along the y-direction. - Initial velocity is zero. ( v_{y0} = 0 ) ##### Upward Throw - Two phases: ascending and descending. - Care about the max Height and Time to reach peak. #### Combined Motion Example: Diver - A diver steps off a 10-meter bridge with an initial horizontal velocity of 1 m/s. - Determine the exact time and speed at impact. #### Trajectory Control - Using gravity, change trajectory of a Projectile has leading factors to control in order to fulfill task Goals. - The angle of release. - The relative height of release. - Apply H (Hierarchical Model) and control horizontal range to find terminal factors. #### Factors Affecting Trajectory - **H Model**: Horizontal range is determined by factors. - Terminal Factors are
\* Speed at Take off
\* Angle Of take off
\* Relative Height #### The Optimal Angle (\theta) - Is dependent on Task goal. - If the goal to minimize Horizontal Range then - The formula to calculate Horizontal Range is: \Delta X = \frac{V_r^2 sin(2\theta)}{G} - Therefore Theta =45 degrees, where Sine is Max! - Same range to hit Horizontal range, The higher the curve, then their will be longer travel - Baseball Pitching, can use low curve, #### Two Angle Possibilities - One angle + another Angle makes up 90 degrees
(\alpha + \beta = 90 Degrees) - Therefore they all have the same
Range=HorizontalRange() #### How To achieve Max Range? - The Angle Theta has to equal 45 degrees - Or around 45! because the more closer is to 45 degrees, the better!
So some javelin Throw or soldier has to be around 45 degrees #### Quick Review of Projectile **The direction of Y is always Vertical!** - There Always change of Velocity With this equation = −9.81 meter per second ( which happens a lot) - Velocity always change.
**The Direction of X is always Horizontal!** - NOVELOCITY ##### Equations we always use - Vertical Equation
VerticalVelocity(tVelocity) = v0 + Integral(tInterval) ##### The Key Point we will always use is - Know your Initial Velocity - It important to know your Vertical or change direction position
So you can take all points during any test and can control it! Because it help you fulfill how to make it What we want to do, which control angle , control height , control speed - Three
Important factor can tell the difference , based on your task goal!