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Motion of a Projectile

Initial Velocity
  • Consider the following diagram
  • This shows that an initial speed U can be defined using vector notations:
    • u = U cos θ i + U sin θ j
  • In this notation:
    • θ defines the angle of elevation from the horizontal
    • i defines the horizontal unit vector
    • j defines the vertical unit vector
Acceleration
  • Gravity affects all projectiles
  • Gravity affects only the vertical component of speed
  • We can thus define acceleration using vector notations:
    • a = 0i – gj = -gj
  • In this notation:
    • g is the acceleration due to gravity
    • i defines the horizontal unit vector
    • j defines the vertical unit vector
  • NOTE: This model assumes that air resistance is negligible
Final Velocity
  • This defines the velocity of the projectile after t seconds of travel
  • Final velocity, according to SUVAT equations, is also dependent on the initial velocity and any acceleration acting on the projectile:
    • v = u + at
  • The above is a scalar equation which can be written as a vector equation:
    • v = u + at
    • This can be further simplified into vector notations by substituting previous equations:
      • v = U cos θ i + U sin θ j – gtj
      • v = U cos θ i + (U sin θ – gt)j
  • Substituting any value of t into the above equation produces the velocity of the projectile at that time
Displacement
  • The distance the particle has moved from the origin
  • Displacement is dependent on initial velocity and acceleration, or final velovity and acceleration:
    • s = ut + ½at²
    • s = vt – ½at²
  • These scalar equations can be written as vector equations:
    • r = ut + ½a
    • r = vt – ½a
  • Substituting previous equations into either of the above equations defines displacement in vector notations
    • r = (U cos θ i + U sin θ j)t – ½gt²j
    • r = U cos θ i + (U sin θ t – ½gt²)j
  • Substituting any value of t into the above equation produces the displacement of the projectile at that time
Special Events
  • Time at Maximum Vertical Height: The time at which the projectile reaches its maximum height, at a given initial velocity and angle of elevation
    • This occurs when vertical component of velocity is 0: U sin θ – gt = 0
    • ∴ TaMVH = U sin θ/g
  • Maximum Vertical Height: is the maximum height the projectile reaches during its flight
    • This occurs when the vertical component of velocity is 0:
    • MVH = U(TaMVH) sin θ – ½g(TaMVH)²
  • Horizontal Range: is the horizontal distance the projectile covers at a given initial velocity and angle of elevation
    • It occurs when the vertical displacement is 0:
      • Ut sin θ – ½gt² = 0
    • Find t and substitute into the equation below:
      • HR = Ut cos θ
  • Maximum Horizontal Range: is the maximum possible horizontal distance the projectile can cover at a given initial velocity
    • It occurs when the angle of elevation is 45°
    • MHR = Ut × √2/2
Examples
  1. A ball was projected at an angle 60° to the horizontal. One second later, another ball was projected from the same point at an angle of 30° to the horizontal. One second after the second ball was released, the two balls collided. Show that the velocities of the balls were 12.99 ms-1 and 15 ms-1. Take the value of g to be 10 ms-2
    • At time zero:
    • At t = 1:
    • At t = 2:
    • Let the velocity of the first ball be U1, and the velocity of the second ball be U2
    • Displacement from origin is equal for both at impact:
      • r1 = r2
      • r1 = (U1(2) cos 60) i + (U1(2) sin 60 – 5(2)²)j
      • r2 = (U2(1) cos 30) i + (U2(1) sin 30 – 5(1)²)j
    • Equate the horizontal components:
      • ∴ 2 U1 cos 60 = U2 cos 30
      • U1 = √3/2 U2
    • Equate vertical components and substitute the information above:
      • ∴ 2 U1 sin 60 – 20 = U2 sin 30 – 5
      • [2(√3/2)(√3/2) – 1/2]U2 = 15
      • ∴ U2 = 15 ms-1
    • Find U1:
      • U1 = √3/2 (15)
      • U1 = 12.99 ms-1
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