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.

Answer 1

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