Projectile Motion Practice Launch at an Angle This topic can be found on pp. 99-101 of your book. “Sacred Rain Arrow” (1998) Allan Houser

Practice Problem #1 Anita Knapp is a world class javelin thrower. If she throws a javelin with a velocity of 28.2 m/s at 42.5o, how far will the javelin go? First you will need to draw a picture. Then, because the initial velocity is at an angle, we need to resolve the velocity vector into its x and y components. We will then need to organize our knowns and unknowns.

Practice Problem #1 Anita Knapp is a world class javelin thrower. If she throws a javelin with a velocity of 28.2 m/s at 42.5o, how far will the javelin go? x component cosθ = 𝑎𝑑𝑗 ℎ𝑦𝑝 cos 42.5° = 𝑣 𝑥 28.2 𝑚/𝑠 vx = 20.8 m/s y component sinθ = 𝑜𝑝𝑝 ℎ𝑦𝑝 sin 42.5° = 𝑣 𝑦 28.2 𝑚/𝑠 viy = 19.1 m/s 28.2 m/s 19.1 m/s 42.5o 20.8 m/s

Practice Problem #1 x-direction y-direction Δx = ? Δy = 0m (why?) Anita Knapp is a world class javelin thrower. If she throws a javelin with a velocity of 28.2 m/s at 42.5o, how far will the javelin go? x-direction y-direction Δx = ? Δy = 0m (why?) t = ? v = 20.8 m/s vi = 19.1 m/s vf = -19.1 m/s (why?) a = -9.81 m/s2 28.2 m/s 19.1 m/s 42.5o 20.8 m/s

Practice Problem #1 x-direction y-direction Δx = ? Δy = 0m t = ? Anita Knapp is a world class javelin thrower. If she throws a javelin with a velocity of 28.2 m/s at 42.5o, how far will the javelin go? x-direction y-direction Δx = ? Δy = 0m t = ? v = 20.8 m/s vi = 19.1 m/s vf = -19.1 m/s a = -9.81 m/s2 Since there is not enough information in the x-direction to solve for Δx, we will need to solve for t in the y-direction first and then use that to solve for Δx.

Practice Problem #1 x-direction y-direction Δx = ? Δy = 0m t = ? Anita Knapp is a world class javelin thrower. If she throws a javelin with a velocity of 28.2 m/s at 42.5o, how far will the javelin go? x-direction y-direction Δx = ? Δy = 0m t = ? v = 20.8 m/s vi = 19.1 m/s vf = -19.1 m/s a = -9.81 m/s2 vf = vi + at -19.1m/s = 19.1m/s + (-9.81m/s2)t t = 3.89 s

Practice Problem #1 x-direction y-direction Δx = ? Δy = 0m t = ? Anita Knapp is a world class javelin thrower. If she throws a javelin with a velocity of 28.2 m/s at 42.5o, how far will the javelin go? x-direction y-direction Δx = ? Δy = 0m t = ? t = 3.89s v = 20.8 m/s vi = 19.1 m/s vf = -19.1 m/s a = -9.81 m/s2 vf = vi + at -19.1m/s = 19.1m/s + (-9.81m/s2)t t = 3.89 s

Practice Problem #1 x-direction y-direction Δx = ? Δy = 0m t = 3.89s Anita Knapp is a world class javelin thrower. If she throws a javelin with a velocity of 28.2 m/s at 42.5o, how far will the javelin go? x-direction y-direction Δx = ? Δy = 0m t = 3.89s v = 20.8 m/s vi = 19.1 m/s vf = -19.1 m/s a = -9.81 m/s2 vf = vi + at -19.1m/s = 19.1m/s + (-9.81m/s2)t t = 3.89 s

Practice Problem #1 x-direction y-direction Δx = ? Δy = 0m t = 3.89s Anita Knapp is a world class javelin thrower. If she throws a javelin with a velocity of 28.2 m/s at 42.5o, how far will the javelin go? x-direction y-direction Δx = ? Δy = 0m t = 3.89s v = 20.8 m/s vi = 19.1 m/s vf = -19.1 m/s a = -9.81 m/s2 𝑣 𝑥 = ∆𝑥 ∆𝑡 20.8 𝑚/𝑠= ∆𝑥 3.89 𝑠 Δx = 80.9 m

Practice Problem #2 A cannon ball is shot off of the edge of a 85.0 m high cliff at an enemy ship. The ball has an initial velocity of 45.2 m/s at 38.1o. How far from the base of the cliff will the cannon ball hit? Again we will first need to draw a diagram and then resolve the velocity vector into its components.

Practice Problem #2 A cannon ball is shot off of the edge of a 85.0 m high cliff at an enemy ship. The ball has an initial velocity of 45.2 m/s at 38.1o. How far from the base of the cliff will the cannon ball hit? x component cosθ = 𝑎𝑑𝑗 ℎ𝑦𝑝 cos 38.1° = 𝑣 𝑥 45.2 𝑚/𝑠 vx = 35.6 m/s y component sinθ = 𝑜𝑝𝑝 ℎ𝑦𝑝 sin 38.1° = 𝑣 𝑦 45.2 𝑚/𝑠 viy = 27.9 m/s 45.2 m/s 27.9 m/s 38.1o 35.6 m/s

Practice Problem #2 x-direction y-direction Δx = ? Δy = -85.0m t = ? A cannon ball is shot off of the edge of a 85.0 m high cliff at an enemy ship. The ball has an initial velocity of 45.2 m/s at 38.1o. How far from the base of the cliff will the cannon ball hit? x-direction y-direction Δx = ? Δy = -85.0m t = ? v = 35.6 m/s vi = 27.9 m/s vf = ? m/s a = -9.81 m/s2 45.2 m/s 27.9 m/s 38.1o 35.6 m/s

Practice Problem #2 A cannon ball is shot off of the edge of a 85.0 m high cliff at an enemy ship. The ball has an initial velocity of 45.2 m/s at 38.1o. How far from the base of the cliff will the cannon ball hit? ∆𝑦= 𝑣 𝑖 𝑡+ 1 2 𝑎 𝑡 2 -85.0m=(27.9m/s)t + ½(-9.81m/s2)t2 t = 7.89 s x-direction y-direction Δx = ? Δy = -85.0m t = ? v = 35.6 m/s vi = 27.9 m/s vf = ? m/s a = -9.81 m/s2

Practice Problem #2 𝑣 𝑥 = ∆𝑥 ∆𝑡 35.6 m/s = ∆𝑥 7.89 𝑠 Δx = 281 m A cannon ball is shot off of the edge of a 85.0 m high cliff at an enemy ship. The ball has an initial velocity of 45.2 m/s at 38.1o. How far from the base of the cliff will the cannon ball hit? 𝑣 𝑥 = ∆𝑥 ∆𝑡 35.6 m/s = ∆𝑥 7.89 𝑠 Δx = 281 m x-direction y-direction Δx = ? Δy = -85.0m t = 7.89 t = 7.89 s v = 35.6 m/s vi = 27.9 m/s vf = ? m/s a = -9.81 m/s2

Practice Problem #3 Answer: 5.34 m above the ground. Stan Still is standing 16.3 m away from a wall. If he throws a water balloon with a velocity of 15.2 m/s at 61.2o at the wall, how high on the wall will the water balloon hit? Answer: 5.34 m above the ground.

y-direction equations Projectile Motion Equations Summary x-direction equation y-direction equations 𝑣= ∆𝑥 ∆𝑡 𝑣 𝑓 = 𝑣 𝑖 +𝑎∆𝑡 ∆𝑥= 1 2 (𝑣 𝑖 + 𝑣 𝑓 )∆𝑡 ∆𝑥= 𝑣 𝑖 ∆𝑡+ 1 2 𝑎 ∆𝑡 2 ∆𝑥= 𝑣 𝑓 ∆𝑡− 1 2 𝑎 ∆𝑡 2 𝑣 𝑓 2= 𝑣 𝑖 2+2𝑎∆𝑥

Important Points vx is always constant in projectile motion constant!!! You must resolve the initial velocity into its x- and y- components. Acceleration is only due to gravity (down) and is always the same value: -9.81 m/s2. The initial direction and angle of the projectile is irrelevant. Time is the only variable that must be the same in both the x and y directions and is therefore the link between the two directions.