# Vectors and Kinematics – (Part 2 of 2)

You can find the first part of this series here: Vectors and Kinematics (Part 1 of 2).

## Projectile Motion

Remember these acronyms: **soh, cah, toa.**

soh = sinº = opposite / hypotenuse

cah = cosº = adjacent / hypotenuse

toa = tanº = opposite / adjacent

- The opposite side is the one that is opposite the angle ‘x’

### Example

Consider a rocket is launched off at a 30º angle to the ground at 10 m/s.

#### Q. How far will it travel? (Horizontal displacement)

S denotes displacement

|| V

_{y}|| denotes magnitude of object in direction ySOH, CAH, TOA

- sin(30) = || V
_{y}|| / 10 m/s - 10 × sin(30) = || V
_{y}|| - 10 × ½ = || V
_{y}|| - 5 m/s = || V
_{y}||`← This is the vertical component`

However, the vertical component goes up, **as well as down.**

Initial = V_{i} = 5 m/s

Final = V_{f} = -5 m/s

- Δv
_{y}= -5 m/s – 5m/s ⇒ -10 m/s - Δv
_{y}= a_{y}× Δt - -10 m/s = -9.8 m/s
^{2}× Δt - Divide both sides by -9.8 m/s
^{2} - ∴ 1.02 seconds = Δt

For the horizontal component:

- Cos(30) = || Vx || / 10 m/s
- 10 × Cos(30) = || Vx ||
- || Vx || = 5√3 m/s

- S
_{x}= 5√3 m/~~s~~× 1.02~~seconds~~ - S
_{x}= 8.83 meters

For the vertical component:

V_{y} = 5 m/s

Y axis meaning the up/down axis.

The positive value (+5 m/s) = the projectile is travelling up (y-axis); or to the right (x-axis). A negative value here would mean it was travelling downward (y-axis), or to the left (x-axis).

So, V_{y} = 5 m/s, meaning the object is travelling up @ 5 m/s

#### Q. How long does the rocket stay in the air?

- S = V
_{average}× Δt - S = (V
_{i}+ V_{f})/2 × Δt

V

_{f}= V_{i}+ a × Δt

Deriving the formula:

- S = (V
_{i}+ V_{i}+ a × Δt)/2 × Δt - S = (V
_{i}+ a × Δt/2) × Δt **S = V**_{i }× Δt + (a +Δt^{2}/2)

Now, at the start, S = 0

- 0 = 5 × Δt – (9.8(Δt
^{2})/2) - 0 = Δt × (5 – 4.9 × Δt) ← one of these underlined, must equal 0
- If we assume the first Δt = 0, then:
- 5 = 4.9 × Δt
- 5/4.9 = (
~~4.9~~× Δt)/~~4.9~~ **Δt = 1.02 seconds**

You must be logged in to post a comment.