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Vectors and projectile motion and worked examples
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Vectors and projectile motion and worked examples

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    Vectors and projectile motion and worked examples Vectors and projectile motion and worked examples Presentation Transcript

    • Stage 2 Physics Section 1Motion in Two Dimensions
    • What is motion?What types of motion are there?What causes motion?How to we describe motion in Physics?
    • What is a vector?Vector quantities have magnitude (size) anddirection.Scalar quantities have magnitude only.Length represents the vectors magnitude.
    • Scalar VectorDistance DisplacementSpeed VelocityMass Acceleration ForceTime weight
    • Velocity vector of an angry bird
    • Resultant of Two VectorsThe resultant is the sum or the combined addition of two vectorquantitiesVectors in the same direction: 6N 4N = 10 N Vectors in opposite directions: 6m 10 m =
    • Vectors• Vectors can be oriented to the gravitational field (up, down or some angle to the horizontal) or compass points (NESW). 5 ms-1 5 ms-1 30o above the horizontal
    • Velocity Vectors• Velocity can be resolved into its horizontal and vertical components at any instant. v vV vH
    • SOHCAHTOA Hypotenuse Opposite Adjacent
    • v vV = v sinvH = v cos
    • Example 1Resolve the following velocity vector into its horizontal and vertical components 30o
    • Example 1Resolve the following velocity vector into its horizontal and vertical components 30o
    • This problem can be solved in two ways(and you need to be able to do both)1. Scale Diagram2. Trigonometry
    • 1. Scale diagramBy drawing a vector diagram (using aprotractor and a ruler) to scale we cansimply measure the size of thecomponents ideally the vector should be10 cm or larger (for accuracy)
    • 2. Trigonometry vvertical = v sin = 40 sin 30o = 20 m s-1 vhorizontal = v cos30o = 40 cos 30o = 34.6 m s-1
    • Example 2Determine the velocity vector with initialhorizontal velocity component of 50 ms –1and vertical 20 ms-1.
    • v=? vv= 20 m s-1 vh= 50 m s- 1
    • 1. Scale diagram Again we could accurately draw the figure and measure the resultant length and angle to find the direction of v.(Note: You need to have a clear Perspex ruler and a protractor for EVERY test and exam)
    • 2. Pythagoras & Trigonometry By Pythagoras theory:v2 = vV2 + vH2 v = v V 2 + vH 2 = 202 + 502 = 2900 = 53.9 m s-1 v=? vv= 20 m s-1 vh= 50 m s- 1
    • tan vV/vHtan = 20/50 = 21.8oie. v = 53.9 m s-1 at 21.8o above the horizontal
    • Summary – Vectors in 2D• Given any vector quantity in 2D it can be resolved into horizontal and vertical components eg displacement, force, fields etc• Given the horizontal & vertical components you can determine magnitude and direction of the vector (formula)
    • Motion in a UniformGravitational Field
    • In the absence of gravity objects move with constant velocity in a straight line. An object will remain at rest, or continue to move at a constant velocity, unless a net force acts on it.Note: The following is all in the absence of air resistance.
    • When an object falls under theinfluence of gravity, the verticalforce causes a constantacceleration
    • vH The resultant motion is a combination of both horizontal and vH verticalvV vV components
    • Horizontal Projection While the vertical component undergoes constantIf an object is acceleration.projectedhorizontally, thehorizontal componentmoves with constantvelocity.
    • Three equations of motionNote that all the equations have “a” in them –they only apply under CONSTANT acceleration
    • Constant vertical acceleration Vertical formulae 2 v 2 v0 2as 1 2Horizontal velocity is constant s vt at 2Horizontal formula v vo at s vH t
    • Three equations of motion
    • Learning Symbols in PhysicsQuantity Quantity Symbol Units Unit symbol
    • ExampleA stone is dropped down a well and takes 3seconds to hit the ground.a) How fast does it hit the bottom?b) How deep is the well?
    • A stone is dropped down a well attakes 3 seconds to hit the ground.a) How fast does it hit the bottom?
    • b) How deep is the well?
    • An arrow is fired upwards at 50ms-1.a) How high does the arrow fly?
    • An arrow is fired upwards at 50ms-1.b) How long does the arrow take to hit theground?
    • Projectile Motion ProblemsExcept for time, everything can be separated into horizontal and vertical components and treated separately. sV = height V0 s H = range t = time of flight
    • Projectile Motion ProblemsHorizontal projection: down is +veUni-level projection: Up is considered positive, and down is negative.(Acceleration due to gravity aV = -9.8ms-2) sV = height V0 s H = range t = time of flight
    • Projectile Motion ProblemsAt the top of the parabolic path, vV= 0 ms-1 1 vV 0ms 2 aV 9.8ms sV height V0 sH range t = time of flight
    • Projectile Motion ProblemsRemember the time of flight is the time it takes to go up+ down. 1 vV 0ms 2 aV 9.8ms sV height V0 sH range t = time of flight
    • Example 1
    • Bi-level projection• An object is projected at a height
    • Maximum RangeTo get the maximum range sH max in a vacuum(no air resistance) the launch angle must be 45o sH max
    • For a projectile launched at ground level find by sample calculation the launch angle that results in a maximum range
    • Pairs of launch angles that yield the same range add up to 90o α + θ = ranges Projectile 90o for various angles of launch 500 450 400 350 300height 250 200 150 100 50 0 0 200 400 600 800 1000 1200 range
    • α + θ = 90oFind the launch angle that yields the same range as 32oθ = 32 α=? α + θ = 90o
    • The Effect of Air ResistanceAir resistance acts in the opposite direction to motion. vertical horizontal
    • The Effect of Air Resistance vertical horizontalThis decreases the• height• rangeSlight decrease in time of flight of the projectile
    • The magnitude of Fair resistance Fair resistance
    • Speed
    • ShapeAerodynamicteardrop
    • SizeMore surface area = more air resistance
    • TextureSmoothRough
    • Air densityLow air density = less air resistanceHigh air density = high air resistance
    • Projectiles in SportConsider the effect of launch height on range
    • As the object has further to fall tflight isincreased.As the object is in the air for longer ittravels farther.
    • 45o 41o For objects at h=0 the optimal angle is 45oFor heights › 0θ max height is less than 45o