Two Dimensional Kinematics Lecture 06: Two Dimensional Kinematics Vector Review 2-D Kinematics Projectile Motion Plenty of time.

2-D Problem Solving X and Y are INDEPENDENT! Break 2-D problem into two 1-D problems. y x

Vectors Position, Velocity, and Acceleration are Vectors! When solving problems, break vectors up into their x- and y-components! (Remember trig…) y x V Vy  Vx

Kinematics in Two Dimensions x = x0 + v0x t + ½ ax t2 vx = v0x + ax t vx2 = v0x2 + 2 ax x y = y0 + v0yt + ½ ay t2 vy = v0y + ay t vy2 = v0y2 + 2 ay y x and y motions are independent but they share a common time t

Projectile Motion ax = 0 ay = -g y = y0 + v0y t - ½ gt2 vy = v0y – g t vy2 = v0y2 – 2 g y x = x0 + v0x t vx = v0x

Summary of Concepts X and Y directions are Independent! Position, velocity and acceleration are vectors Consider each direction separately Projectile Motion ax = 0 in horizontal direction ay = -g in vertical direction

2-D Kinematics Example A cannon ball is shot from off a cliff at a speed of 68 m/s and an angle of 35º above the horizontal. If the cliff is 85 m off the ground, how far from the base of the cliff does the cannon ball land? y x

2-D Kinematics Example A cannon ball is shot from off a cliff at a speed of 68 m/s and an angle of 35º above the horizontal. If the cliff is 85 m off the ground, how far from the base of the cliff does the cannon ball land? We are looking for the HORIZONTAL position when the ball lands, so eventually we will need to use: x = x0 + v0x t x0 (the initial x-position) we will define as 0 m v0x we can get from trig: (68 m/s)*(cos 35º) t will need to come from somewhere else…

2-D Kinematics Example t = 9.74 s A cannon ball is shot from off a cliff at a speed of 68 m/s and an angle of 35º above the horizontal. If the cliff is 85 m off the ground, how far from the base of the cliff does the cannon ball land? To find the time before the ball hits the ground we will use the VERTICAL direction: y = y0 + v0y t – ½ g t2 y = 0 m and y0 = 85 m v0y we can get from trig: (68 m/s)*(sin 35º) g = 9.8 m/s2 so now we use the quadratic formula to find t: t = 9.74 s

2-D Kinematics Example x = 543 m A cannon ball is shot from off a cliff at a speed of 68 m/s and an angle of 35º above the horizontal. If the cliff is 85 m off the ground, how far from the base of the cliff does the cannon ball land? Finally, to get HORIZONTAL position use: x = x0 + v0x t x0 (the initial x-position) we defined as 0 m v0x we calculated from trig: (68 m/s)*(cos 35º) = 55.7 m/s t = 9.74 s x = 543 m

Equilibrium x direction: SF = m a -TL cosq + TR cosq = 0 TL = TR .12 m Determine the tension in the 6 meter rope if it sags 0.12 meters in the center when a gymnast with weight 250 Newtons is standing on it. y x x direction: SF = m a -TL cosq + TR cosq = 0 TL = TR 3 m .12 m q y direction: SF = m a TL sinq + TR sinq - W = 0 2 T sinq = W T = W/(2 sinq) = 3115 N 10

2-D Kinematics Example A cannon is shot from off a cliff at a speed of 68 m/s and an angle of 35º above the horizontal. If the cliff is 85 m off the ground, how far from the base of the cliff does the cannon ball land? y x