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a high diver of mass 58.3 kg steps off a board 10.0 m above the water and falls vertical to the water, starting from rest. if her downward motion is stopped 1.90 s after her feet first touch the water, what average upward force did the water exert on her?

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The average upward force exerted by the water on the diver is 572.97 N.

Initial velocity (v_initial): 0 m/s (starts from rest)

Final velocity (v_final): 0 m/s (stopped by water)

Acceleration due to gravity (g): 9.81 m/s^2

Height (h): 10.0 m

Mass of the diver (m): 58.3 kg

Time taken to stop (t): 1.90 s

Work done by the water:

W = F * h

Change in kinetic energy:

KE = (1/2) * m * (v_final^2 - v_initial^2)

Setting work equal to change in energy:

F * h = (1/2) * m * (v_final^2 - v_initial^2)

Solving for the upward force (F):

F = [(1/2) * m * (v_final^2 - v_initial^2)] / h

Calculating the final velocity (v_final):

v_final = v_initial + g * t

v_final = 0 + 9.81 * 1.90 = 18.66 m/s

Substituting values to find F:

F = [(1/2) * 58.3 * (18.66^2 - 0^2)] / 10.0

F = 572.97 N

Therefore, the correct upward force exerted by the water on the diver is 572.97 N.

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