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Projectile Motion Diver Jump Question

  • Thread starter lvn
  • Start date
  • #1
lvn
2
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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?
 
Last edited:

Answers and Replies

  • #2
lvn
2
0
Figured it out.
 

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