A diver springs upward from a board that is three meters above the water. At the instant she contacts the water her speed is 9.10 m/s and her body makes an angle of 80.0° with respect to the horizontal surface of the water. Determine her initial velocity, both magnitude and direction.

To determine the diver's initial velocity, we need to break down the given information and apply some trigonometry.

Let's assume the x-direction is parallel to the horizontal surface of the water, and the y-direction is perpendicular to it.

Given:
Height of the board (h) = 3 meters
Final speed (v) = 9.10 m/s
Angle with respect to the horizontal surface (θ) = 80.0°

Now let's find the initial velocity components in the x- and y-directions using the following equations:

v₀x = v * cos(θ)
v₀y = v * sin(θ)

Step 1: Calculate the velocity component in the x-direction (v₀x):
v₀x = 9.10 m/s * cos(80.0°)
v₀x = 9.10 m/s * 0.173648 (cosine of 80.0°)
v₀x = 1.58 m/s (rounded to two decimal places)

Step 2: Calculate the velocity component in the y-direction (v₀y):
v₀y = 9.10 m/s * sin(80.0°)
v₀y = 9.10 m/s * 0.984808 (sine of 80.0°)
v₀y = 8.844 m/s (rounded to three decimal places)

Thus, we have the initial velocity components:
v₀x = 1.58 m/s (velocity in the x-direction)
v₀y = 8.844 m/s (velocity in the y-direction)

To determine the magnitude and direction of the initial velocity, we can use the Pythagorean theorem and inverse tangent function.

Step 3: Calculate the magnitude of the initial velocity (v₀):
v₀ = √(v₀x² + v₀y²)
v₀ = √(1.58 m/s)² + (8.844 m/s)²
v₀ ≈ 9.03 m/s (rounded to two decimal places)

Step 4: Calculate the direction of the initial velocity (θ₀):
θ₀ = tan⁻¹(v₀y / v₀x)
θ₀ = tan⁻¹(8.844 m/s / 1.58 m/s)
θ₀ ≈ 80.8° (rounded to one decimal place)

Therefore, the diver's initial velocity has a magnitude of approximately 9.03 m/s and makes an angle of approximately 80.8° with respect to the horizontal surface of the water.