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 71.0° with respect to the horizontal surface of the water. Determine her initial velocity, both magnitude and direction.

To determine the initial velocity of the diver, we need to analyze the motion of the diver's body in both the horizontal and vertical directions.

1. Vertical Motion:
Let's consider the vertical direction as the positive direction. We can start by finding the time it takes for the diver to reach the water's surface using the vertical motion equation:
v = u + gt

Here,
v = final vertical velocity = 0 m/s (as the diver comes to rest at the water's surface)
u = initial vertical velocity
g = acceleration due to gravity = 9.8 m/s²
t = time of flight

Since the diver springs upward, the initial velocity in the vertical direction (u) will be positive. Therefore, we can rewrite the equation as:
0 = u + (9.8 m/s²)t

Now, let's calculate the time of flight (t) by rearranging the equation:
t = -u / (9.8 m/s²)

2. Horizontal Motion:
Based on the information provided, we know that the diver's body makes an angle of 71.0° with respect to the horizontal surface of the water. This means the horizontal component of the initial velocity can be calculated using the trigonometric function cosine.
cos(θ) = adjacent / hypotenuse

Here,
θ = 71.0°
adjacent = horizontal component of initial velocity (u_x)
hypotenuse = initial velocity (u)

Rearranging the equation, we get:
u_x = u * cos(θ)

3. Finding u and u_x:
Now, let's solve the equations to find the values of u and u_x.

Substituting the value of t from equation (1) into equation (2), we have:
u_x = u * cos(θ) = -u / 9.8 * cos(71.0°)

Given that the speed of the diver just before hitting the water is 9.00 m/s, we can write:
u² = u_x² + u_y² (using Pythagorean theorem)

Substituting the value of u_x and considering u = 9.00 m/s, we can solve for u_y:
u_y = √(9.00² - u_x²)

4. Final Calculation:
Now, we can substitute the values of u_x and u_y back into the initial velocity equation to find the magnitude and direction of the initial velocity (u).

u = √(u_x² + u_y²)
direction = arctan(u_y/u_x)

Evaluate the calculations and substitute the values.

To determine the initial velocity of the diver, we can analyze the vertical and horizontal components separately.

1. Vertical Component:

The vertical displacement of the diver can be calculated using the formula:
Δy = Vyi * t + (1/2) * g * t^2

Since the diver starts at rest in the vertical direction when she contacts the water, ΣVfy = 0. This gives us:
0 = Vyi - g * t

Solving the above equation for Vyi, we find:
Vyi = g * t

Given that g = 9.8 m/s^2 and the diver contacted the water with a speed of 9.00 m/s, we can substitute these values into the equation to find t:
9.00 m/s = 9.8 m/s^2 * t
t = 9.00 m/s / 9.8 m/s^2
t ≈ 0.918 sec

Now, we can substitute the value of t back into the equation for Vyi to find the vertical component of the initial velocity:
Vyi = 9.8 m/s^2 * 0.918 sec
Vyi ≈ 8.99 m/s (upward)

2. Horizontal Component:

The horizontal displacement of the diver can be calculated using the formula:
Δx = Vxi * t

Since the horizontal speed remains constant, Vxi = Vxf, where Vxf is the horizontal component of the final velocity, which is also the horizontal component of the initial velocity.

Given that the speed at which the diver contacts the water is 9.00 m/s, we can use trigonometry to find the horizontal component of the initial velocity:
Vxf = Vf * cosθ
Vxf = 9.00 m/s * cos(71.0°)
Vxf ≈ 3.42 m/s

Now that we have the vertical and horizontal components of the initial velocity, we can use the Pythagorean theorem to find the magnitude of the initial velocity:
Vinitial = sqrt(Vxi^2 + Vyi^2)
Vinitial = sqrt((3.42 m/s)^2 + (8.99 m/s)^2)
Vinitial ≈ 9.47 m/s

To find the direction of the initial velocity, we can use the inverse tangent function:
θ = atan(Vyi / Vxi)
θ = atan(8.99 m/s / 3.42 m/s)
θ ≈ 69.4° (above the horizontal)

Therefore, the initial velocity of the diver is approximately 9.47 m/s at an angle of 69.4° (above the horizontal).