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Unformatted text preview: 6.25. Model: Assume the particle model for the projectile and motion in a plane. Visualize: Solve: (a) Using ( 29 ( 29 2 1 2 2 2 2 y y y y v t t a t t = + + , ( 29 ( 29 ( 29 ( 29 2 2 1 2 2 0 m 30 m/s sin60 7.5 s 0 s 9.8 m/s 7.5 s 0 s 80.8 m y = + ° + =  Thus the launch point is 80.8 m higher than where the projectile hits the ground. (b) Using ( 29 2 2 1 1 2 y y y v v a y y = + , ( 29 ( 29 ( 29 2 2 2 2 1 1 0 m /s 30sin60 m/s 2 9.8 m/s 0 m 34.4 m y y = ° + ⇒ = (c) The x component is ( 29 2 cos60 30 m/s cos60 15 m/s x x v v v = = ° = ° = . The ycomponent is ( 29 ( 29 2 2 2 1 sin60 y y y v v a t t v g t t = + = °  ( 29 ( 29 ( 29 2 30 m/s sin60 9.8 m/s 7.5 s 0 s = °  47.52 m/s =  ( 29 ( 29 2 2 15 m/s 47.52 m/s 49.8 m/s v ⇒ = +  = 2 1 1 2 47.52 tan tan 72.5 15 y x v v θ = = = ° below + x Assess: An angle of 72.5 ° made with the ground, as the projectile hits the ground 80.8 m below its launch point, is reasonable in view of the fact that the projectile was launched at an angle of 60 ° . 6.27. Model: The golf ball is a particle following projectile motion. Visualize: (a) The distance traveled is x 1 = v x t 1 = v cos θ × t 1 . The flight time is found from the yequation, using the fact that the ball starts and ends at y = 0: 2 1 1 1...
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This homework help was uploaded on 04/17/2008 for the course PHYS 2014 taught by Professor Nandi during the Spring '08 term at Oklahoma State.
 Spring '08
 Nandi
 Physics

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