Physics, 19.01.2021 18:50 katrinanuez

A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 3.00 m/s, and her takeoff point is 1.40 m above the pool. (a) How long are her feet in the air?
s
(b) What is her highest point above the board?
m
(c) What is her velocity when her feet hit the water?
m/s

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Answer from: Paris4life

a) t= 0.92 s

b) h = 0.46 m

c) v = -6.04 m/s

Explanation:

In order to find the total time that the feet are in the air, we must add two times:1) time needed to reach to the maximum height (t₁) 2) time from when starts to fall from the maximum height until her feet hit the water (t₂)In order to get t₁, we need to take into account that at her highest point, the vertical speed will be zero.Taking for granted the value for the acceleration due to gravity,g = -9.8 m/s2, we can apply the definition of acceleration, and replacing by the givens, we can find t₁ as follows:

$t_{1} = \frac{v_{o}}{g} = \frac{3.0m/s}{9.8m/s2} = 0.3 s (1)$

In order to find t₂, we need to find first the highest point above the board, which is indeed what is asked for in b).We can use the following kinematic equation, taking into account that at the highest point, the final velocity vf will be zero.The equation can be written as follows:

$v_{f} ^{2} - v_{o} ^{2} = 2*g*\Delta h (2)$

Replacing by the givens, and solving for Δh, we get:

$\Delta h = \frac{v_{o}^{2}}{2*g} = \frac{(3.0m/s)^{2} }{2*9.8m/s2} = 0.46 m (3)$

The total height over the water will be just the sum of the takeoff point (1.40 m over the water) and the value we found in (3):H = 1.40 m + 0.46 m = 1.86 mNow, we can use the equation that relates the vertical displacement with the time, remembering that v₀=0, as follows:

$H = \frac{1}{2}*g*t^{2} (4)$

Replacing by the givens and the value found for H in (4), and solving for t, we get the value of t₂, as follows:

$t_{2} = \sqrt{\frac{2*H}{g} } = \sqrt{\frac{2*1.86m}{9.8m/s2} } = 0.62 s (5)$

The total time will the sum of t₁ and t₂:t = 0.3 s + 0.62 s = 0.92 s (6)

As we have just found, the highest point above the board was at 0.46m above the takeoff point, so the highest point above the board is just 0.46 m.

In order to find the velocity when her feet hit the water, we can use the same equation (2), taking into account that v₀=0, and Δh = H= -1.86m.Solving (2) for vf, we get:

$v_{f} = - \sqrt{2*g*H} = -\sqrt{2*9.8m/s2*1.86m} = -6.04 m/s (7)$

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