Question

A diver springs upward from a board that is 2.89 meters above the water. At the instant she contacts the water her speed is 9.04 m/s and her body makes an angle of 75.1° with respect to the horizontal surface of the water. Determine her initial velocity.

Answer 1

Horizontal component, vx = 9.04 cos75.1 = 2.32 m/s

Vertical component, vy = 9.04 sin75.1 = 8.74 m/s

Vertical component of initial velocity,

uv = sqrt(vy^2 - 2as) = sqrt(8.74^2 - (2 x 9.8 x 2.89))

uv = 4.44 m/s

Horizontal component of initial velocity, ux = 2.32 m/s

Resultant, u = sqrt(2.32^2 + 4.44^2) = 5.01 m/s

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