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.
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/s
Then 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)
Vo = 11.25m/s
Theta:
Tan^-1 (Voy/Vox)
= 96
Answer was wrong - what am I doing wrong?
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/s
Then 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)
Vo = 11.25m/s
Theta:
Tan^-1 (Voy/Vox)
= 96
Answer was wrong - what am I doing wrong?
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