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2014_SP08_Sol04 - 6.25 Model Assume the particle model for...

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6.25. Model: Assume the particle model for the projectile and motion in a plane. Visualize: Solve: (a) Using ( 29 ( 29 2 1 2 0 0 2 0 2 0 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 0 1 0 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 0 0 cos60 30 m/s cos60 15 m/s x x v v v = = ° = ° = . The y -component is ( 29 ( 29 2 0 2 0 0 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 ° .
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6.27. Model: The golf ball is a particle following projectile motion. Visualize: (a) The distance traveled is x 1 = v 0 x t 1 = v 0 cos θ × t 1 . The flight time is found from the y -equation, using the fact that the ball starts and ends at y = 0: 2 0 1 1 1 0 0 1 1 0 1 1 1 2 2 2 sin 0 sin ( sin ) v y y v t gt v gt t t g θ θ θ - = = - = - = Thus the distance traveled is 2 0 0 1 0 2 sin 2
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