1. Projectile Motion Pt. 1
Week

2. Student will: Objective
Analyze and describe accelerated motion in 2D using horizontal projectile motion

3. Cornell Notes (3/3) Launch horizontally means Launch Angle is 0°
Velocity in x-dir ALWAYS the same
Acceleration in x-dir ALWAYS 0
Velocity in y-dir ALWAYS changes
Acceleration in y-dir ALWAYS -9.8 m/s2

4. Time is equal in both equations!
Optional
Since projectile motion is 2D, you need TWO sets of kinematic equations
Set for x-dir
Set for y-dir
∆𝑥= 𝑣 𝑖,𝑥 + 𝑣 𝑓,𝑥 ∆𝑡
𝑣 𝑓,𝑥 = 𝑣 𝑖,𝑥 + 𝑎 𝑥 ∆𝑡
∆𝑥= 𝑣 𝑖,𝑥 ∆𝑡 𝑎 𝑥 ∆𝑡 2
𝑣 𝑓,𝑥 2 = 𝑣 𝑖,𝑥 2 +2 𝑎 𝑥 ∆𝑥
∆𝑦= 𝑣 𝑖,𝑦 + 𝑣 𝑓,𝑦 ∆𝑡
𝑣 𝑓,𝑦 = 𝑣 𝑖,𝑦 + 𝑎 𝑦 ∆𝑡
∆𝑦= 𝑣 𝑖,𝑦 ∆𝑡 𝑎 𝑦 ∆𝑡 2
𝑣 𝑓,𝑦 2 = 𝑣 𝑖,𝑦 2 +2 𝑎 𝑦 ∆𝑦
Time is equal in both equations!

5. Cornell Notes (1/5)
Example: Horizontal Projectile Motion One Direction (all of them) are shove off a cliff with an initial horizontal velocity of 65 m/s and lands 195 meters away from the base of the cliff. How tall is the cliff?

6. Cornell Notes (2/5) 𝑎 x =0 𝑚 s 2 ∆ 𝑥 =195 m 𝑣 i,𝑦 =0 𝑚 𝑠
Steps
Define
Given: Unknown: Dy = ?
x-direction
𝑣 i,x =65 𝑚 𝑠
𝑎 x =0 𝑚 s 2
∆ 𝑥 =195 m
y-direction
𝑣 i,𝑦 =0 𝑚 𝑠
𝑎 y =−9.81 𝑚 s 2
One Direction (all of them) are shove off a cliff with an initial horizontal velocity of 65 m/s and lands 195 meters away from the base of the cliff. How tall is the cliff?

7. Vectors

8. Cornell Notes (3/5) Δ x = v i,x Δt Δt= Δ x v i,x
Plan
Choose an equation or situation:
Rearrange the equation to isolate the unknown:
∆𝑥= 𝑣 𝑖,𝑥 + 𝑣 𝑓,𝑥 ∆𝑡
𝑣 𝑓,𝑥 = 𝑣 𝑖,𝑥 + 𝑎 𝑥 ∆𝑡
∆𝑥= 𝑣 𝑖,𝑥 ∆𝑡 𝑎 𝑥 ∆𝑡 2
𝑣 𝑓,𝑥 2 = 𝑣 𝑖,𝑥 2 +2 𝑎 𝑥 ∆𝑥
Δ 𝑥 = 𝑣 𝑖,𝑥 Δ𝑡 𝑎 𝑥 ∆𝑡 2
Δ 𝑦 = 𝑣 𝑖,𝑦 ∆𝑡 𝑎 𝑦 ( Δ𝑡) 2
Δ x = v i,x Δt a x ∆t 2
∆𝑦= 𝑣 𝑖,𝑦 + 𝑣 𝑓,𝑦 ∆𝑡
𝑣 𝑓,𝑦 = 𝑣 𝑖,𝑦 + 𝑎 𝑦 ∆𝑡
∆𝑦= 𝑣 𝑖,𝑦 ∆𝑡 𝑎 𝑦 ∆𝑡 2
𝑣 𝑓,𝑦 2 = 𝑣 𝑖,𝑦 2 +2 𝑎 𝑦 ∆𝑦
Δ y = 𝑣 𝑖,𝑦 ∆t a y ( Δt) 2
Δ x = v i,x Δt
Δt= Δ x v i,x
Δ y = a y ( Δt) 2

9. Cornell Notes (4/5) Δ𝑡= Δ 𝑥 𝑣 𝑖,𝑥 Δ𝑡= 195 65 Δ𝑡=3 𝑠𝑒𝑐
Calculate
Substitute the values into the equation and solve
Δ𝑡= Δ 𝑥 𝑣 𝑖,𝑥
Δ𝑡=
Time is equal in both equations!
Δ𝑡=3 𝑠𝑒𝑐
Δ 𝑦 = 𝑎 𝑦 ( Δ𝑡) 2
Plug time into second equation to find distance
Δ 𝑦 = 1 2 −9.8 ( 3) 2
Δ 𝑦 =−44.145
Δ 𝑦 = 𝑚

10. Cornell Notes (5/5)
Evaluate
The answer appears (-) because gravity is (-). But we know height is (+), so we change the (-) to a (+) sign.
The calculated height of the cliff is m tall.