Microsoft PowerPoint - Knight Ch 4 - kinematics in TWo Dimendions - new

Microsoft PowerPoint - Knight Ch 4 - kinematics in TWo Dimendions - new

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Unformatted text preview: 1 i f Δ r r r − = t Δ Δ = r v dt d t lim t r r v = Δ Δ = → Δ Kinematics in Two Dimensions –Chapter 4 ____________ j ˆ v i ˆ v j ˆ dt dy i ˆ dt dx dt r d v y x + = + = = r r j ˆ y i ˆ x r + = r Position and velocity are expressed in terms of both horizontal and ____________components. 2D motion with Constant Acceleration ____________ ____________ Relating Position to Velocity 2 y 2 x y x v v v sin v dt dy v cos v dt dx v + = θ = = θ = = ____________ ____________ Acceleration Vector dt d t lim t v v a = Δ Δ = → Δ 2 Acceleration Vectors j dt dv i dt dv a y x ˆ ˆ + = r ____________ Projectile (2D) Motion A ____________ is any body that is given an initial velocity and then follows a path determined entirely by the effects of __________________ acceleration and _________resistance. The path followed by a projectile is called its _________ . In projectile motion, the ________ acceleration is due to gravity! This means there is ________ acceleration in the “x”direction! Therefore, x and y velocity components are ___________ of one another! Projectile (2D) Motion What does this mean mathematically? a x = _________ a y = _________ Projectile motion, no air resistance 6.14 Projectile motion equations t x Δt v x x ix i f − + = For a projectile with motion in the x and y direction, () t y Δt g 2 1 Δt v y y 2 iy i f − − + = t v const v v ix fx − = = t v gΔΔ v v iy fy − − = 3 x y CAR BALL cosΘ gt 2 1 t v x x 2 ox o + + = cosΘ gt 2 1 t v x x 2 ox o + + = y = sinΘ gt 2 1 t v y 2 oy − = The equations in x are _____________, therefore, when the ball returns to the x axis, it will ___________________________. θ ____________ ____________ ______ Maximum Height What is the maximum height y (v=0, t=?) attained by the projectile? At the maximum height sin = − = gt v v i y θ Plug t maxH into position equation 2 2 1 sin gt t v y y i i − + = θ g v t i H θ sin max = g v H i 2 sin 2 2 θ = 2 sin 2 1 sin sin − + = g v g g v v H i i i θ θ θ ____________ _____ ________ Maximum Range What is the maximum range x(t = t tot ) attained by the projectile? Plug t TOTAL into position equation θ θ cos sin 2 = g v v R i i t TOTAL = _____________ Identity: 2 cos θ sin θ = sin 2 θ θ cos t v x x i i + = g v R i θ 2 sin 2 = ________ Playing Darts Playing darts, one player throws a dart towards the bulls eye located 8.00 ft (2.44 m) away. If the velocity of the dart when it leaves his hands parallel to the floor is 5.00 m/s, how far below the bulls eye will it hit the target? v o x y 2 2 2 1 sin 2 1 cos t a t v y y t a t v x x y i i x i i + + = + + = θ θ 2 y ox t a 2 1 y and t v x = = Solving the x-t equation for t and plug into y-t equation solving for y 2 2 1 = = ix y ix v x a y and v x t ( ) m s m m s m y 17 . 1 / 00 . 5 44 . 2 / 80 . 9 2 1 2 2 − = − = ____ ____________...
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This note was uploaded on 04/17/2008 for the course PHYSICS 122 taught by Professor Pope during the Spring '08 term at Clemson.

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Microsoft PowerPoint - Knight Ch 4 - kinematics in TWo Dimendions - new

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