Projectile Motion Pt. 1 Week

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

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

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= 1 2 𝑣 𝑖,𝑥 + 𝑣 𝑓,𝑥 ∆𝑡 𝑣 𝑓,𝑥 = 𝑣 𝑖,𝑥 + 𝑎 𝑥 ∆𝑡 ∆𝑥= 𝑣 𝑖,𝑥 ∆𝑡+ 1 2 𝑎 𝑥 ∆𝑡 2 𝑣 𝑓,𝑥 2 = 𝑣 𝑖,𝑥 2 +2 𝑎 𝑥 ∆𝑥 ∆𝑦= 1 2 𝑣 𝑖,𝑦 + 𝑣 𝑓,𝑦 ∆𝑡 𝑣 𝑓,𝑦 = 𝑣 𝑖,𝑦 + 𝑎 𝑦 ∆𝑡 ∆𝑦= 𝑣 𝑖,𝑦 ∆𝑡+ 1 2 𝑎 𝑦 ∆𝑡 2 𝑣 𝑓,𝑦 2 = 𝑣 𝑖,𝑦 2 +2 𝑎 𝑦 ∆𝑦 Time is equal in both equations!

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?

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?

Vectors

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= 1 2 𝑣 𝑖,𝑥 + 𝑣 𝑓,𝑥 ∆𝑡 𝑣 𝑓,𝑥 = 𝑣 𝑖,𝑥 + 𝑎 𝑥 ∆𝑡 ∆𝑥= 𝑣 𝑖,𝑥 ∆𝑡+ 1 2 𝑎 𝑥 ∆𝑡 2 𝑣 𝑓,𝑥 2 = 𝑣 𝑖,𝑥 2 +2 𝑎 𝑥 ∆𝑥 Δ 𝑥 = 𝑣 𝑖,𝑥 Δ𝑡+ 1 2 𝑎 𝑥 ∆𝑡 2 Δ 𝑦 = 𝑣 𝑖,𝑦 ∆𝑡+ 1 2 𝑎 𝑦 ( Δ𝑡) 2 Δ x = v i,x Δt+ 1 2 a x ∆t 2 ∆𝑦= 1 2 𝑣 𝑖,𝑦 + 𝑣 𝑓,𝑦 ∆𝑡 𝑣 𝑓,𝑦 = 𝑣 𝑖,𝑦 + 𝑎 𝑦 ∆𝑡 ∆𝑦= 𝑣 𝑖,𝑦 ∆𝑡+ 1 2 𝑎 𝑦 ∆𝑡 2 𝑣 𝑓,𝑦 2 = 𝑣 𝑖,𝑦 2 +2 𝑎 𝑦 ∆𝑦 Δ y = 𝑣 𝑖,𝑦 ∆t+ 1 2 a y ( Δt) 2 Δ x = v i,x Δt Δt= Δ x v i,x Δ y = 1 2 a y ( Δt) 2

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= 195 65 Time is equal in both equations! Δ𝑡=3 𝑠𝑒𝑐 Δ 𝑦 = 1 2 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 =44.145 𝑚

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 44.145 m tall.