#### Motion of a Projectile

__Initial Velocity__

__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__

__Acceleration__

- Gravity affects all projectiles
- Gravity affects only the vertical component of speed
- We can thus define acceleration using vector notations:
**a**= 0**i**– g**j**= -g**j**

- 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__

__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**+**a**t- This can be further simplified into vector notations by substituting previous equations:
**v**= U cos θ**i**+ U sin θ**j**– gt**j****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__

__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**=**u**t + ½**a**t²**r**=**v**t – ½**a**t²

- Substituting previous equations into either of the above equations defines displacement in vector notations
**r**= (U cos θ**i**+ U sin θ**j**)t – ½gt²**j****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__

__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__

__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 U
_{1}, and the velocity of the second ball be U_{2} - Displacement from origin is equal for both at impact:
- ∴
**r**_{1}=**r**_{2} **r**_{1}= (U_{1}(2) cos 60)**i**+ (U_{1}(2) sin 60 – 5(2)²)**j****r**_{2}= (U_{2}(1) cos 30)**i**+ (U_{2}(1) sin 30 – 5(1)²)**j**

- ∴
- Equate the horizontal components:
- ∴ 2 U
_{1}cos 60 = U_{2}cos 30 - U
_{1}=^{√3}/_{2}U_{2}

- ∴ 2 U
- Equate vertical components and substitute the information above:
- ∴ 2 U
_{1}sin 60 – 20 = U_{2}sin 30 – 5 - [2(
^{√3}/_{2})(^{√3}/_{2}) –^{1}/_{2}]U_{2}= 15 - ∴ U
_{2}= 15 ms^{-1}

- ∴ 2 U
- Find U
_{1}:- U
_{1}=^{√3}/_{2 }(15) - U
_{1}= 12.99 ms^{-1}

- U

- At time zero: