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A diver springs upward from a board that is 3.00 m above the water. At the instant she contacts the water her speed is 8.90 m/s...Asked by Kenisha Frinks
A diver springs upward from a board that is 2.90 meters above the water. At the instant she contacts the water her speed is 9.11 m/s and her body makes an angle of 78.2° with respect to the horizontal surface of the water. Determine her initial velocity.
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Answered by
bobpursley
vhorizontal=9.11cos78.2
vf=9.11Sin78.2
figure both of those.
In the vertical:
vf^2=vi^2-2ah=vi^2-2*9.8*2.90
solve for vi, the initial vertical velocity.
finally, the initial velocity is make up of horizontal and vertical components...
v=sqrt(vi^2+vh^2) where vi is the initial vertical velocity, and vh is the initial horizontal velocty
vf=9.11Sin78.2
figure both of those.
In the vertical:
vf^2=vi^2-2ah=vi^2-2*9.8*2.90
solve for vi, the initial vertical velocity.
finally, the initial velocity is make up of horizontal and vertical components...
v=sqrt(vi^2+vh^2) where vi is the initial vertical velocity, and vh is the initial horizontal velocty
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