A diver springs upward from a board that is three meters above the water. At the instant she contacts the water her speed is 9.30 m/s and her body makes an angle of 83.0° 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 break down the given information into horizontal and vertical components.
First, let's find the vertical component of the initial velocity. We know that the diver springs upward, so the vertical component of her velocity is responsible for this motion. The vertical component can be calculated using the equation:
v_y = v * sin(angle)
where v_y is the vertical component of the initial velocity, v is the speed of the diver at contact with the water (9.30 m/s), and the angle is 83.0°.
Plugging in the values:
v_y = 9.30 m/s * sin(83.0°) ≈ 9.30 m/s * 0.988 ≈ 9.18 m/s
Now, let's find the horizontal component of the initial velocity. Since the diver is springing vertically upward, the horizontal component does not contribute to the overall motion. Therefore, the horizontal component of the initial velocity is zero.
So, the initial velocity, both in magnitude and direction, is:
Magnitude: v = √(v_x^2 + v_y^2)
= √(0 + 9.18 m/s)^2
≈ 9.18 m/s
Direction: The direction of the initial velocity is given by the tangent of the angle between the initial velocity vector and the horizontal surface of the water. Since the horizontal component is zero, the direction is purely vertical, pointing upward.
Therefore, the initial velocity of the diver is approximately 9.18 m/s in the upward direction.