Projectile Motion

Projectile Motion Objects that are thrown are called projectiles. The path of a projectile is a curve called a parabola.

Projectile motion ∆ 𝑑 𝑦 = 𝑣 𝑖𝑦 𝑡− 1 2 𝑔 𝑡 2 ∆ 𝑑 𝑥 = 𝑣 𝑖𝑥 𝑡 A projectile’s motion can be broken down into the horizontal and vertical components: ∆ 𝑑 𝑦 = 𝑣 𝑖𝑦 𝑡− 1 2 𝑔 𝑡 2 ∆ 𝑑 𝑥 = 𝑣 𝑖𝑥 𝑡 Key Concepts: Time is the same in both the x and y Acceleration is gravity- 9.81 m/s2 Down is negative *Remember to break the initial velocity into its components!

Launched at an angle Vertical and Horizontal components to initial velocity: 𝑣 𝑖𝑥 = 𝑣 𝑖 𝑐𝑜𝑠𝜃 𝑣 𝑖𝑦 = 𝑣 𝑖 𝑠𝑖𝑛𝜃 12 m/s 40° 𝑣 𝑖𝑥 =12𝑐𝑜𝑠40 𝑣 𝑖𝑦 =12𝑠𝑖𝑛40

Example Mark drives his motorcycle off a horizontal ramp at 325 m/hr and lands a horizontal distance of 30 m away from the edge of the ramp. What is the height of the ramp? Variables: Δ 𝑑 𝑥 = 30 m Δ 𝑑 𝑦 = ? a = -9.81 m/s2 𝑣 𝑥 =325 𝑚/ℎ𝑟

Example ∆ 𝑑 𝑦 = 𝑣 𝑦 𝑡+ 1 2 𝑎 𝑡 2 ∆ 𝑑 𝑥 = 𝑣 𝑥 𝑡 Variables: a = -9.81 m/s2 𝑣 𝑥 =325 𝑚/ℎ𝑟 30 = (325)t t = 30/325 t = 0.09 sec ∆ 𝑑 𝑦 = (0)(0.09) + ½(-9.81)( 0.09 2 ) ∆ 𝑑 𝑦 = - 0.04 m

Example A baseball is thrown at an angle of 25° at a speed of 23 m/s. If the ball was caught 42 m from the thrower, how long was it in the air? How high did the ball travel? Variables: Θ = 25 t= ? V = 23 𝑑 𝑦 = ? 𝑑 𝑥 = 42

Example ∆ 𝑑 𝑦 = 𝑣 𝑦 𝑡 − 1 2 𝑎 𝑡 2 ∆ 𝑑 𝑥 = 𝑣 𝑥 𝑡 Variables: ∆ 𝑑 𝑦 = 𝑣 𝑦 𝑡 − 1 2 𝑎 𝑡 2 ∆ 𝑑 𝑥 = 𝑣 𝑥 𝑡 Example Variables: Θ = 25 t= ? 𝑣 𝑖𝑥 = 𝑣 𝑖 𝑐𝑜𝑠𝜃 V = 23 𝑑 𝑦 = ? 𝑣 𝑖𝑥 =23𝑐𝑜𝑠25 𝑑 𝑥 = 42 𝑣 𝑖𝑥 = 20.8 42 = (20.8)t t = 42/20.8 t = 2.01sec ∆ 𝑑 𝑦 = 9.72(2.01) - ½(-9.81)( 2.01 2 ) ∆ 𝑑 𝑦 = 39.4 m

Finding 𝑣 𝑖 𝑣 𝑖 = 2 −𝑔 𝑑 𝑥 2𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃 𝑣 𝑖 = 2 −𝑔 𝑑 𝑥 2𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃 What is the initial velocity of an object thrown a horizontal distance of 250 m at an angle of 30°?