Projectiles at an Angle IB Physics | Unit 2 | Motion

Reminder of our Equations Units m m s-1 m s-2 s 𝑣=𝑢+𝑎𝑡 𝑢 𝑣 𝑎 𝑡 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 𝑠 𝑣 2 = 𝑢 2 +2𝑎𝑠 𝑠= 𝑣+𝑢 𝑡 2

2-D Problem Solving Steps Start with “suvat” in the vertical direction and pretend it’s just a freefall problem The air time is the same for horizontal motion Solve for horizontal using v = s/t Vertical Only 𝑠 𝑢 𝑣 𝑎 𝑡

Remember Vectors? vx vy v = 24 m s-1 vy 55° vx

One Dimensional Motion

Horizontal Projectile Vertical Only 𝑠 𝑢 𝑣 𝑎 𝑡

Two Dimensional Projectile

Projectile – First Half vx = 1st Half Vertical 𝑠 𝑢 𝑣 𝑎 𝑡 vy = 24 m s-1 55° 24 m s-1 19.7 m s-1 55° 13.8 m s-1

Projectile – Full Thing 𝑇𝑜𝑡𝑎𝑙 𝑇𝑖𝑚𝑒 𝐷𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑥 24 m s-1 55° 24 m s-1 19.7 m s-1 55° 13.8 m s-1

Projectile – In General vx = 1st Half Vertical 𝑠 𝑢 𝑣 𝑎 𝑡 vy = u uy θ ux 𝑇𝑜𝑡𝑎𝑙 𝑇𝑖𝑚𝑒 𝐷𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑥

Try This… 𝑠 𝑢 𝑣 𝑎 𝑡 1st Half Vertical vx = 30 m s-1 vy = 24° You hit a baseball at 24° above the horizontal as a speed of 30 m s-1. How far does the ball travel before it hits the ground?