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)
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
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
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