Week

Projectile Motion Pt. 1

Objective

Student will: Analyze and describe accelerated

motion in 2D using horizontal projectile motion

Cornell Notes (3/3)

Velocity in x-dir ALWAYS the same

Velocity in y-dir ALWAYS changes

Launch horizontally means Launch Angle

is 0

Acceleration in x-dir ALWAYS 0

Acceleration in y-dir ALWAYS -9.8 m/s2

OptionalSince projectile motion is 2D, you need TWO sets of kinematic equations

Set for x-dir Set for y-dir

Time is equal in both equations!

Cornell Notes (1/5)

Example: Horizontal Projectile MotionOne 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?

Cornell Notes (2/5)Given:

Unknown: Dy = ?

Steps1) Define

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?

y-direction

x-direction

Vectors

Cornell Notes (3/5)Choose an equation or situation:

Rearrange the equation to isolate the unknown:

2) Plan

Δ �⃗�=�⃗� 𝑖 ,𝑥❑ Δ𝑡+

12�⃗�𝑥 (∆𝑡 )2

Δ �⃗�=�⃗� 𝑖 , 𝑦∆ 𝑡+12�⃗�𝑦 ¿

Δ t=Δ x⃗v⃗ i , x

❑

Δ y⃗=12a⃗ y¿

Δ x⃗= v⃗ i , x❑ Δt+

12a⃗x (∆ t )2

Δ y⃗=�⃗� 𝑖 ,𝑦∆ t+12a⃗y ¿

Cornell Notes (4/5)Substitute the values into the equation and solve

3) Calculate

Δ �⃗�=12�⃗�𝑦¿

Δ 𝑡=Δ �⃗��⃗�𝑖 ,𝑥

❑

Plug time into second equation to find distance

Δ 𝑡=3𝑠𝑒𝑐

Δ �⃗�=44.145𝑚Δ �⃗�=−44.145

Time is equal in both equations!

Δ 𝑡=19565

Δ �⃗�=12

(−9.8 )¿

Cornell Notes (5/5)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 44.145 m tall.

4) Evaluate