Projectile Motion & Forces

Projectile Motion

Basic Concepts

  • Kinematic Equations: These equations are used for each component of motion (x and y). Time is the only common variable between them.
  • 2D & 3D Kinematic Equations:
    • x-component: x<em>f=12a</em>xt2+v<em>i,xt+x</em>ix<em>f = \frac{1}{2} a</em>x t^2 + v<em>{i,x} t + x</em>i
    • y-component: y<em>f=12a</em>yt2+v<em>i,yt+y</em>iy<em>f = \frac{1}{2} a</em>y t^2 + v<em>{i,y} t + y</em>i
    • x-velocity: v<em>f,x=a</em>xt+vi,xv<em>{f,x} = a</em>x t + v_{i,x}
    • y-velocity: v<em>f,y=a</em>yt+vi,yv<em>{f,y} = a</em>y t + v_{i,y}
    • x-velocity squared: v<em>f,x2=v</em>i,x2+2axΔxv<em>{f,x}^2 = v</em>{i,x}^2 + 2 a_x \Delta x
    • y-velocity squared: v<em>f,y2=v</em>i,y2+2ayΔyv<em>{f,y}^2 = v</em>{i,y}^2 + 2 a_y \Delta y
    • Vector form: r<em>f=12at2+v</em>it+ri\vec{r}<em>f = \frac{1}{2} \vec{a} t^2 + \vec{v}</em>i t + \vec{r}_i
    • Velocity vector: v<em>f=at+v</em>i\vec{v}<em>f = \vec{a} t + \vec{v}</em>i

Clicker Question 9

  • Question: If a ball is released from rest and another is shot horizontally from the same height, which hits the ground first?
  • Answer: C) Both hit at the same time.
  • Explanation: Vertical motion is independent of horizontal motion. Video explanation at: https://www.youtube.com/watch?v=zMF4CD7i3hg (at 1:00)

Gravitational Field

  • Direction: Downward.
  • Magnitude: g=9.8ms2g = 9.8 \frac{m}{s^2} near Earth’s surface.
  • Decreases with distance from Earth’s center.
  • gg is four times smaller if twice as far from the center.

Projectile Motion Details

  • Basic Idea: Free fall with horizontal motion.
  • Assumptions:
    • Negligible air resistance.
    • Not too fast or light.
    • Uniform gravitational field, g\vec{g}.
    • Near Earth's surface.
    • Small trajectories compared to Earth.
  • g9.8ms2\vec{g} \approx 9.8 \frac{m}{s^2}

Projectile Motion Tools

  • Acceleration components:
    • a=g\vec{a} = \vec{g}
    • ax=0a_x = 0
    • ay=ga_y = -g
    • g=9.8ms2g = 9.8 \frac{m}{s^2}
  • 2D kinematic equations with constant acceleration during free fall.

Trajectory

  • The trajectory of projectile motion is a downward parabola.

Problem Solving Steps

  1. Diagram the problem:
    • Include the trajectory.
    • Identify time intervals of constant acceleration.
    • Label moments where you know information.
    • Label moments where you want to know information.
  2. List Knowns and Unknowns: At each significant moment.
  3. Decide which kinematic equations will convert knowns into desired unknowns.

Problem Solving Tips

  • Time relates horizontal and vertical motion (it's the only common variable).
  • Unstated Information:
    • Max height implies vy=0v_y = 0
    • Known acceleration in projectile motion.
  • Be careful with timing:
    • Differentiate between just before vs just after landing.
    • Velocity after hitting the ground is usually irrelevant.

Example Problem: Fire Fighting

  • Scenario: A firefighter needs to spray water into a 1 m tall window 12 m above the ground. The nozzle is 1 m above the ground and 20 m from the building. If the water exits the nozzle at 35.0° above the horizontal at a speed of 30 m/s, will it enter the window?
  • Solution: No, the water hits at 11.8 m above the ground, which is 0.2 m below the window.

Class Problem: Is it a Homerun?

  • Scenario: A baseball is hit at a 40° angle from a height of 1 m above the ground. There is a 4 m tall wall 100 m away. Ignore air resistance.
  • Question: How fast was the baseball moving right after it was hit to barely clear the wall?
  • Answer: 32.1 m/s (approximately 72 mph).

Horizontal Range

  • Requirements:
    • Same start and end height (level ground).
    • Free fall.
  • Formula: R=v<em>i2sin2θ</em>igR = \frac{v<em>i^2 \sin 2\theta</em>i}{g}
    • RR = Range
    • viv_i = Initial velocity
    • θi\theta_i = Launch angle from horizontal
    • gg = Gravitational field magnitude

Maximum Height

  • Basic Idea: How high the projectile reaches above launch height.
  • Requirements: Free fall and level ground.
  • Formula: y<em>max=v</em>i2sin2θi2gy<em>{max} = \frac{v</em>i^2 \sin^2 \theta_i}{2g}
    • ymaxy_{max} = Max height
    • viv_i = Initial velocity
    • θi\theta_i = Launch angle from horizontal
    • gg = Gravitational field magnitude

Clicker Question 11

  • Question: On Earth, the longest recorded golf drive was 471 m. About how far would it have gone on the moon, where gravity is 6 times weaker?
  • Answer: C) 471(6) m

Example Problem: The Long Jump

  • Scenario: A daredevil barely jumps over 24 trucks (each 3 m wide) off a 30° ramp.
  • Question: How fast was he going when he left the ramp?
  • Answer: 28.5 m/s or 63.8 mph.

Class Problem: Cannonball!

  • Scenario: A cannon is fired at an angle of 60° across level ground. Neglecting air resistance, the cannonball leaves the barrel at 90 m/s.
  • Questions:
    • How high does the cannonball reach?
    • How far does the cannonball go before hitting the ground?
  • Answers:
    • Height: 310 m
    • Distance: 716 m

Circular Motion

Basic Concepts

  • Easier to use directions relative to the circle's center.
  • Definitions:
    • Centripetal (cc): Toward the center.
    • Tangential (tt): Counterclockwise along the circle.
  • Use cc & tt for components like xx and yy.

Circular Motion: Constant Speed

  • Change in velocity is toward the center over short times.
  • Acceleration is toward the center = “Centripetal” component, aca_c.
  • Formula: ac=v2ra_c = \frac{v^2}{r}
    • aca_c = Centripetal acceleration
    • vv = Speed
    • rr = Radius of the circle

Circular Motion: Changing Speed

  • Turning requires centripetal acceleration, aca_c (how quickly velocity is turning).
  • Changing speed requires tangential acceleration, ata_t (how quickly speed is changing).
  • Direction:
    • Speeding up: a\vec{a} toward v\vec{v} (< 90°).
    • Slowing down: a\vec{a} opposite v\vec{v} (> 90°).
  • Formulas:
    • ac=v2ra_c = \frac{v^2}{r}
    • a<em>t=dv</em>tdta<em>t = \frac{dv</em>t}{dt} (slope of tangent line on tangential component vtv_t of velocity vs. time graph).

Circular Motion Variables

  • Period, TT: Time for a full rotation.
  • Frequency, f=1Tf = \frac{1}{T}: Number of cycles per second.
  • Angular Speed, ω=2πf\omega = 2\pi f: Number of radians turned per second.
  • These apply to constant speed cases.

Speed & Acceleration From Angular Variables

  • Speed: v=2πrT=2πfr=rωv = \frac{2\pi r}{T} = 2\pi f r = r \omega
  • Centripetal Acceleration: ac=v2r=4π2rT2=4π2f2r=ω2ra_c = \frac{v^2}{r} = \frac{4\pi^2 r}{T^2} = 4\pi^2 f^2 r = \omega^2 r
  • Requirements: Circular motion and constant speed.
  • rr = radius of circle, TT = period, ff = frequency, ω\omega = angular velocity

Forces

Introduction to Forces

  • Rough Definition: Push and pull.
  • Technical Definition: Interactions between two physical objects that pass the