A 65.0-kg woman steps off a 11.0-m diving platform and drops straight down into the water. 1) If she reaches a depth of 4.60 m, what is the average resistance force exerted on her by the water

2160.26 N.

Explanation:

From the question,

The total energy of the woman = work done by the water.

mg(h+d) = F'×d Equation 1

m = mass of the the woman, g= acceleration due to gravity, h = height of the platform, d = depth of water reached by the woman, F'= Resistance force exerted on her by the water.

make F' the subject of the equation

F' = mg(h+d)/d Equation 2

Given: m = 65.0 kg, g = 9.8 m/s², h = 11 m, d = 4.6 m.

Substitute into equation 2

F' = 65(9.8)(11+4.6)/4.6

F' = 9937.2/4.6

F' = 2160.26 N.

real, smaller and upside-down.   so the answer should be b.