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Projectile Motion Diver Jump Question
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[QUOTE="lvn, post: 4519551, member: 489475"] [b]A diver springs upward from a board that is 3.60 m above the water. At the instant she contacts the water her speed is 13.1 m/s and her body makes an angle of 67.3 ° with respect to the horizontal surface of the water. Determine her initial velocity, both (a) magnitude and (b) direction.[/b] Basically what I've tried is I found initial velocity component X by finding the final velocity X component: Vx = Vox Vx = 13.1cos67.3 Vox = 5.06m/sThen I found the initial velocity Y component through the kinematic equation: Vy^2 = Voy^2 - 2a(y-yo) Voy = sqrt(171.61 - 2(-9.8)(-3.6)) Voy = sqrt(171.61 - -19.6(-3.6)) Voy = sqrt(171.61 - 70.56) Voy = sqrt(101.05) Voy = 10.05m/s Using pythagoras to find Vo: Vo = sqrt(10.05^2 + 5.06^2) Vo = sqrt(101.0025 + 25.6036) [b]Vo = 11.25m/s[/b] Theta: Tan^-1 (Voy/Vox) [b]= 96[/b] Answer was wrong - what am I doing wrong? [/QUOTE]
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