 4 Ijesha Close, Ilupeju, Lagos
+2347 086 296 002

#### 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
Minimum 4 characters