Question

A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and her takeoff point is 1.80 m above the pool. What is her velocity when her feet hit the water?

Answer 1

7.16 m/s

Step-by-step explanation:

The swimmer’s initial velocity is 4.00 m/s and her takeoff point is 1.80 m above the pool 1. We can use the kinematic equation that relates the final velocity (vf), initial velocity (vi), acceleration (a), and displacement (d) to find the swimmer’s velocity when her feet hit the water:

vf^2 = vi^2 + 2ad

Since the swimmer is falling straight down, we can take the displacement to be equal to the distance between her takeoff point and the water, which is 1.80 m 1. We can also take acceleration due to gravity to be -9.81 m/s^2 1. Substituting these values into the equation, we get:

vf^2 = (4.00 m/s)^2 + 2(-9.81 m/s^2)(-1.80 m) vf^2 = 16.00 m2/s2 + 35.29 m2/s2 vf^2 = 51.29 m2/s2

Taking the square root of both sides, we get:

vf = sqrt(51.29) m/s vf ≈ 7.16 m/s

Therefore, the swimmer’s velocity when her feet hit the water is approximately 7.16 m/s.