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tugas fisika teknik kimia

Jun 02, 2018

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    CHAPTER 3

    PROJECTILE MOTION

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    Projectile motion:a body in free fall moving intwo dimensions that is subject to the forces of

    gravity and air resistance and its path is aparabola

    Bodies projected into the air areprojectiles.

    Examples of moving body as projectile:

    Shot putt High jumper Long jumper

    Soccer ball Baseball Ski jumper

    What are the components of the movement?

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    The speed in the x-direction

    is constant; in the y-

    direction the object moveswith constant acceleration

    g.

    This photograph shows two ballsthat start to fall at the same time.

    The one on the right has an initial

    speed in the x-direction. It can be

    seen that vertical positions of the

    two balls are identical at identical

    times, while the horizontal

    position of the yellow ball

    increases linearly.

    y

    x

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    If an object is launchedat an initial angle of 0

    with the horizontal, the analysis is similar except

    that the initial velocity has a verticalcomponent.This diagram isdrawn using a

    correct scale.

    Which velocity ischanging as time

    goes by?

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    Objectives when launching projectiles:

    Influence time of flight

    Maximum

    Minimum

    Maximize horizontal displacement

    Maximize vertical displacement

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    Factors Influencing Projectile Trajectory

    Angle of projection

    Projection speed

    Relative height of projection

    Trajectory: the flight path of a projectile

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    General shapes

    Perfectly vertical (angle = ?)

    Parabolic (angle = ?)

    Perfectly horizontal (angle = ?)

    Implications in sports

    Air resistance may causeirregularities

    Angle of Projection

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    Range: horizontal displacement.

    For oblique projection angles,

    speed determines height and

    range

    For vertical projection angle,

    speed determines height.

    Projection speed:

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    Difference between projection

    and landing height

    Greater the relative projection

    height, longer the flight time,

    greater the displacement.

    Relative Projection Height

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    Optimum Projection Conditions

    Maximize the speed of projection Maximize release height

    Optimum angle of projection Release height = 0, then angle = 450

    Release height, then angle

    Release height, then angle

    Complimentary angles (A + B = 90) have the

    same range Large angle higher and slower time: tennis lob,punt

    Small anglelower and faster time: baseball throw

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    Analyzing Projectile Motion

    Initial velocity:

    Horizontal component is constant

    Horizontal acceleration = 0

    Vertical component is constantly changing

    Vertical acceleration = -9.81 m/s2

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    A projectile launched with velocity will have both vertical &

    horizontal components of that velocity.

    Horizontal & Vertical Components:

    Vertical is influenced by gravity

    No force (neglecting air resistance) affects the horizontal

    Horizontal relates to distance

    Vertical relates to maximum height achieved

    Horizontal and vertical components are independent

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    Major influence of verticalcomponent

    Not the horizontal component

    Force of Gravity:

    Constant, unchanging

    Negative acceleration (-9.81

    m/s2)

    Apex:

    Highest Point

    Vertical velocity = 0 m/s

    A body projected

    straight upward will

    have the same speedat the end of its flight

    as it did when it was

    launched

    The Influence of Gravity

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    In a vacuum, horizontal speed of a projectile

    remains constant

    Air resistance affects the horizontal speed of a

    projectile

    This class , ho rizontal veloci ty w i l l be regarded asconstant

    Influence of Air Resistance

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    Equations of

    Constant Acceleration

    Galileos Laws of constant acceleration:

    v2= v1+ at

    S = v1t + at2

    V22= v21+ 2 aS

    S = displacement; v = velocity; a = acceleration; t = time

    Subscript 1 & 2 represent first or initial and second or final

    point in time

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    Equations of

    Constant Acceleration

    Horizontal component : a = 0

    v2= v1S = v1t

    V22= v21

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    Equations of

    Constant Acceleration

    Vertical component: a = -9.81 m/s2, initial velocity of zero

    (a dropped object)

    v2= at

    S = at2

    V22= 2aS

    Vertical component at apex: a = 0

    0 = v21+ 2aS (det. max height)

    0 = v1+ at

    (total flight timemultiply by 2)

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    Solving Problems Involving Projectile Motion

    Projectile motion is motion with constantacceleration in two dimensions, where the

    acceleration is gand is down.

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    Solving Problems Involving Projectile Motion:

    Procedures

    1. Read the problem carefully, and choose theobject(s) you are going to analyze.

    2. Drawa diagram.

    3. Choosean origin and a coordinate system.

    4. Decideon the timeinterval;

    this is the same in both directions, and

    includes only the time the object is movingwith constant acceleration g.

    5. Examinethe xand ymotions separately.

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    Solving Problems Involving Projectile Motion:

    Procedures

    6. List known and unknown quantities.

    Remember that vxnever changes, and that vy=

    0 at the highest point.

    7. Plan how you will proceed. Use the

    appropriate equations; you may have to

    combine some of them.

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    A movie stunt driver on a motorcycle speeds

    horizontally off a 50.0-m-high cliff. How fast

    must the motorcycle leave the cliff top to land

    on level ground below, 90.0 m from the base of

    the cliff where the cameras are? Ignore air

    resistance.

    Driving Off A Cliff

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    A football is kicked at an angle 0 = 37.0 with a

    velocity of 20.0 m/s, as shown. Calculate (a) the

    maximum height, (b) the time of travel before the

    football hits the ground, (c) how far away it hits the

    ground, (d) the velocity vector at the maximum height,

    and (e) the acceleration vector at maximum height.

    Assume the ball leaves the foot at ground level, and

    ignore air resistance and rotation of the ball.

    A Kicked Football

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    A child sits upright in a wagon

    which is moving to the right at

    constant speed as shown. The

    child extends her hand and throwsan apple straight upward (from her

    own point of view), while the

    wagon continues to travel forward

    at constant speed. If air resistanceis neglected, will the apple land (a)

    behind the wagon, (b) in the

    wagon, or (c) in front of the

    wagon?

    Conceptual Problem:

    Where does the apple land?

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    Conceptual Problem: The wrong strategy.

    A boy on a small hill aims his water-balloon slingshothorizontally, straight at a second boy hanging from a

    tree branch a distance daway. At the instant the water

    balloon is released, the second boy lets go and falls

    from the tree, hoping to avoid being hit. Show that he

    made the wrong move. (He hadnt studied physicsyet.) Ignore air resistance.