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 + ½at²
- r = vt – ½at²
- 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
- It occurs when the vertical displacement is 0:
- 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
- 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
- At time zero: