A diver is performing on the 10 meter board at a Olympic style pool (A board tha

Question

A diver is performing on the 10 meter board at a Olympic style pool (A board that is 10 m above the water surface). She runs along the board a 4.5 meters per second and exits the end. On her way to the water she does a half gainer followed by one and a half summersaults. What is her speed when hitting the water? On arriving head first at the water, the diver maneuvers and finally comes to a stop in a horizontal orientation touching the bottom of the 3 meter deep diving circle of the pool, having traveled 5 meters since entering the water surface. What was the force stopping the diver after her entry? Assume the diver has a mass of 63 Kilograms, and is 1.7 meters tall. Also assume that her center of mass is half way between her feet and the top of her head.

Explanation / Answer

let vf be the final velocity and vi be the initial velocity; g = +9.81 m/s^2 and y = +10 meters (vertically downward convention) vf^2 = vi^2 + 2gy Since she didn't have any vertical component of her velocity, vi = 0 vf^2 = 2gd = 2 (9.81 m/s^2) (10 m) = 196.2 m^2/s^2 vf = 14 m/s (vertically downwards) Therefore, her speed components are: x ->vx = 4.5 m/s y ->vy = 14 m/s, and then by Pythagorean Theorem, v = sqrt(x^2 + y^2) = sqrt( 4.5^2 + 14^2) = 14.7 m/s (speed when hitting the water) Now, let's assume that she traveled the 5 meters in the water in a straight line and her total vertical displacement in the water was 3 meters. By using Pythagorean Theorem to determine her total horizontal displacement in the water. x = sqrt(5^2 - 3^2) = 4 meters. We will have a second assumption that at the moment that she reached the bottom of the 3-meter deep pool, she totally had zero acceleration. That means that she was slowed down by a constant force oriented with an angle above the horizontal (this is because she had vertical and horizontal velocities when she entered the water region. Accounting for the x component of the force: Let the x component of the force be A. Since she didn't have any force induced with respect to the horizontal, we can safely assume that the force that stopped her was the only force that was applied to her. Let's compute for her rate of deceleration. Her total traveled distance was 4 meters and her initial speed was 4.5 meters. (vf)^2 = (vi)^2 + 2ad a = -(vi)^2/2d = -(4.5 m/s)^2 / (2)(4 m) = -2.5 m/s^2 (rate of deceleration wrt horizontal) Fx = ma (newton's 2nd law) Fx = 63 (-2.5) = -160 N (horizontal component of the force) Accounting for the vertical component of the force: Therefore, there will be a downward force that will be counteracted by an upward force. Since the diver slowed down before stopping, there would be an an additional upward force that was applied while disregarding the downward acceleration. This can be computed by using the displacement of 3 meters and initial velocity (downwards) of 14 m/s. (vf)^2 = (vi)^2 + 2ad where f.speed = 0, therefore, a = -(vi)^2 / 2d = -(14 m/s)^2 / (2)(3 m) = -33 m/s^2 (upward direction) To compute for the real force applied upwards, we will add a 9.81 m/s^2 acceleration (a counter acceleration to the 9.81 m/s^2 gravitational acceleration to this 33 m/s^2 upward acceleration yielding a total of upward acceleration applied of 33 + 9.81 = 43 m/s^2 (will be negative because we take the upward acceleration to be negative) This means that the total upward force will be (let B be the total upward force): Fy = ma = (63 kg)(-43 m/s^2) = -2709 N Since Fx = -160 N and Fy = -2709 N,

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