A swimmer bounces straight up from a diving board and falls feet first into a po

Question

A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 3.45 m/s, and her takeoff point is 1.61 m above the pool. Assume she remains perfectly straight without bending or rotating, and ignore air resistance.

Draw a picture of this problem first before trying to work it out, and identify the knowns and unknowns.

(a) What is the highest distance of her feet above the water?

    m

(b) How long are her feet in the air?

    s

(c) What is her velocity as her feet pass the board going down? (Use - for the downward direction.)

    m/s

(d) What is her velocity when her feet hit the water?

    m/s

Explanation / Answer

given voy = 3.45 m/s

ay = -g = -9.8 m/s^2

yo = 1.61 m

(a)

at the highest point vy = 0

from equations of motion

2*ay*y - yo = vy^2-voy^2

-2*9.8*(y-1.61) = 0-3.45^2

y = 2.21 m <<<-------answer

(b)

the total displacement dy = -1.61 m

from the equations of motion

dy = voy*T + 0.5*ay*T^2

-1.61 = 3.45*T - 0.5*9.8*T^2

T = 1.02 s <<------answer

++++

c)

as her feet passses the board point displacemet = dy = 0

2*ay*dy = vy^2-voy^2

vy = -voy

vy = -3.45 m/s <<<------------answer

+++++

d)

vy = voy + ay*T

vy = 3.45-(9.8*1.02)

vy = -6.55 m/s <<<------------answer

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