A diver springs upward from a board that is 3.20 m above the water. At the instant she contacts the water her speed is 11.4 m/s and her body makes an angle of 68.7 ° with respect to the horizontal surface of the water. Determine her initial velocity, both (a) magnitude and (b) direction

To determine the initial velocity of the diver, we can break it down into horizontal and vertical components.

(a) To find the horizontal component of the initial velocity, we can use the formula:

Vx = V * cos(θ)

Where:
Vx is the horizontal component of the velocity
V is the total velocity (11.4 m/s in this case)
θ is the angle the body makes with the horizontal (68.7 ° in this case)

Vx = 11.4 m/s * cos(68.7 °)

Calculating the value of Vx gives us:

Vx = 11.4 m/s * cos(68.7 °) ≈ 4.86 m/s

So, the horizontal component of the initial velocity is approximately 4.86 m/s.

(b) To find the vertical component of the initial velocity, we can use the formula:

Vy = V * sin(θ)

Where:
Vy is the vertical component of the velocity
V is the total velocity (11.4 m/s in this case)
θ is the angle the body makes with the horizontal (68.7 ° in this case)

Vy = 11.4 m/s * sin(68.7 °)

Calculating the value of Vy gives us:

Vy = 11.4 m/s * sin(68.7 °) ≈ 10.14 m/s

So, the vertical component of the initial velocity is approximately 10.14 m/s.

To determine the magnitude of the initial velocity, we can use the Pythagorean theorem:

V = sqrt(Vx^2 + Vy^2)

Where:
V is the total velocity (11.4 m/s in this case)
Vx is the horizontal component of the velocity (4.86 m/s in this case)
Vy is the vertical component of the velocity (10.14 m/s in this case)

V = sqrt((4.86 m/s)^2 + (10.14 m/s)^2)

Calculating the value of V gives us:

V ≈ sqrt(23.63 m^2/s^2 + 102.92 m^2/s^2) ≈ sqrt(126.55 m^2/s^2) ≈ 11.25 m/s

So, the magnitude of the initial velocity is approximately 11.25 m/s.

The direction of the initial velocity can be determined by finding the angle it makes with the horizontal. We can use the inverse tangent (arctan) function:

θ = arctan(Vy / Vx)

Where:
θ is the angle the initial velocity makes with the horizontal
Vx is the horizontal component of the velocity (4.86 m/s in this case)
Vy is the vertical component of the velocity (10.14 m/s in this case)

θ = arctan(10.14 m/s / 4.86 m/s)

Calculating the value of θ gives us:

θ ≈ arctan(2.09)

θ ≈ 63.7 °

So, the direction of the initial velocity is approximately 63.7 ° above the horizontal.