Can you help with this physics problem?

Question

A diver springs upward from a board that is three meters above the water. At the instant she contacts the water her speed is 9.00 m/s and her body makes an angle of 82.0° with respect to the horizontal surface of the water. Determine her initial velocity, both magnitude and direction.

Please explain in as much detail as possible.

Answer 1

vf = 9

angle = 82

h0 = 3

vf = sqrt(vx^2+vy^2) = 9

and from that you can find the x and y components of final velocity

vf cos 82 = vx = 1.25m/s

vf sin 82 = vy = 8.91m/s

Recall this common equation: v2^2 = v1^2 + 2ax

vy^2 = vy0^2 - 2gh

vx^2 = vx0^2 => vx0 = vx = 1.25 m/s

you can also find the height at the top of her jump:

2g(h-3)=vy0^2 ..  (it is h-3 due to how the jumping point is 3 meters from the point of impact on the pool water surface)

solve for h:

plug this into the vy equation

vy^2 = (2g(h-3))^2 - 2gh, and plug in h to find vy