A projectile is any object that moves through Projectile Motion! A projectile is any object that moves through the air or through space, acted on only by gravity (and air resistance).

Projectile motion is divided into two sections: Objects launched horizontally Objects launched at angles

HORIZONTAL PROJECTION Motion in which something is launched horizontally. X component remains Constant. Vertical Motion of a Projectile launched Horizontally Horizontal Motion

HORIZONTAL PROJECTION Horizontal velocity is constant. Vertical velocity is changing due to gravitational acceleration..

When solving projectile motion problems it is important to separate motion along the x axis and y axis. Time is universal, it will be the same for both x and y. X Y Vox= Voy = Vx = Vy = ΔX = ΔY= a = a= t = t=

y = -1/2at2 y = -1/2(9.8)(5)2 y = -1/2(9.8)(5)2 y = -122.5 m Example A ball is thrown horizontally at a speed of 30 m/s from the top of a cliff. If the ball hits the ground 5.0 seconds later, approximately how high is the cliff? Vo= 30 m/s y = -1/2at2 g y y = -1/2(9.8)(5)2 y = -1/2(9.8)(5)2 t= 5s y = -122.5 m y = 122.5 m

Take out a separate sheet of paper. We are going to do #3 together!

Horizontal Projectile Recap: We need to split motion into x-component and y-component With Horizontal Projectiles, the Vi in only in the x direction and the acceleration is only in the y-direction. We can simplify the formulas: Displacement Equations: ΔX = Vi,xt + ½ axt2 ΔY = Vi,yt + ½ ayt2 ax = 0 Vi,y = 0 ΔX = Vi,xt ΔY = ½ ayt2 Time can help you solve for a missing variable!

Horizontal Projectile Recap: Horizontal speed stays constant. Vertical velocity will change due to gravity. Velocity Equations: Vi,x = Constant Vf,y = Vi,y + ayt Vf,y2 = Vi,y2 + 2ayΔy Vf,y = ayt Vf,y2 = 2ayΔy There is no Vi,y

Why is this important? What are the real world applications?

PROJECTILES AT AN ANGLE The more general case of projectile motion occurs when the projectile is fired at an angle.

PROJECTILES AT AN ANGLE At what angle will the projectile will travel the furthest? At what angle will the projectile stay in the air longer? 75° (at 90° you get max time in the air!) 45°

PROJECTILES AT AN ANGLE The horizontal velocity component remains the same size throughout the entire motion of the cannonball.

Projectiles Launched at Angles Recap: We need to split motion into x-component and y-component Vy Vx = Vo cosθ Vy = Vo sinθ 10 m/s Vx = 10 cos(30) Vy = 10 sin(30) 30° Vx Vx = 8.66 m/s Vy = 5 m/s ΔX = Vi,xt + ½ axt2 ΔY = Vi,yt + ½ ayt2 ax = 0 Time can help you solve for a missing variable!

Projectiles launched at anglesRecap: Horizontal speed stays constant. Vertical velocity will change due to gravity. First, break the initial velocity into components. Vx = Vo cosθ Vy = Vo sinθ Velocity Equations: Vi,x = Constant Vf,y = Vi,y + ayt Vf,y2 = Vi,y2 + 2ayΔy

Assumptions you CAN make. Acceleration in the y- direction will be -9.8 m/s2 At maximum height Vy= 0 m/s Velocity in the x direction in constant Air resistance is ignored (unless otherwise stated) If a projectile is launched at an angle, it has Vx and Vy If a projectile is launched horizontally, Vy=0

Vx = Vo cosθ Vy = Vo sinθ Vx = 200 cos40 Vy = 200 sin40 X Y Vix= 153.21 m/s Viy= 128.56m/s ax= 0 m/s2 ay= -9.8 m/s2 Vx = 200 cos40 Vy = 200 sin40 Δx= Δy= Vfx= 153.21m/s Vfy= -128.56m/s t= t= Vx = 153.21 m/s Vy = 128.56 m/s 200 m/s Vy 40° Vx -Vfy = Viy

Vfy = Viy + at -128.56 = 0 + (-9.8)t 26.24= t -Vfy = Viy 200 m/s X Y Vix= 153.21 m/s Viy= 128.56m/s ax= 0 m/s2 ay= -9.8 m/s2 -128.56 = 0 + (-9.8)t Δx= Δy= 26.24= t Vfx= 153.21 m/s Vfy= -128.56m/s 26.24s t= t= 26.24s 200 m/s Vy 40° Vx -Vfy = Viy

Vix = Δx t Δx = Vix *t Δx = 153.21 *26.24 Δx = 4020.23 -Vfy = Viy X Y Vix= 153.21 Viy= 128.56 ax= 0 ay= -9.8 Δx = Vix *t Δx= 4020.23 Δy= Vfx= 153.21 Vfy= -128.56 Δx = 153.21 *26.24 t= 26.24s t= 26.24s Δx = 4020.23 200 m/s Vy 40° Vx -Vfy = Viy

Vfy2 = Viy2 + 2aΔy 02 = 128.562 + 2(-9.8)Δy -128.562 = 2(-9.8)Δy X Y Vfy2 = Viy2 + 2aΔy Vix= 153.21 Viy= 128.56 02 = 128.562 + 2(-9.8)Δy ax= 0 ay= -9.8 Δx= 4020.23 Δy= 843.24 -128.562 = 2(-9.8)Δy Vfx= 153.21 Vfy= -128.56 843.24= Δy t= 26.24s t= 26.24s Vy@ max height = 0 200 m/s Vy 40° Vx -Vfy = Viy