A person jumps out of an airplane above the surface of the Earth, and falls a distance h before opening their parachute. Once the prachute is open the person coasts to the ground a distance d at constant velocity. a. The work done on the person by the Earth is:
b. The change in gravitational potential energy of the person + Earch system is:

a)   W_total = mg (2h + d)   , b)     E_total = - mg (h + d)

Explanation:

a) We must solve this problem in two parts, the first for the accelerated movement and the second for the movement with constant speed

Let's look for work for the part that is in free fall

        y = y₀ + v₀ t - ½ g t²

when he jumps out of a plane his vertical speed is zero

        y =y₀ - ½ g t²

        dy = 0 - ½ g 2t dt

the work in this first part is

        W₁ = ∫ F dy

        W₁ = mg ∫ g t dt

        W₁ = m g² t² / 2

the time it takes to travel the distance y₀-y = h is

         y₀-y = ½ g t²

         t =

we substitute

          W₁ = m g² 2h / g

          W₁ = m g 2h

now we look for the work for the part with constant speed

since the velocity is constant let's use the uniform motion ratio

          W₂ = F d

           W₂ = mg d

the total work is

           W_total = W₁ + W₂

           W_total = 2mgh + m gd

           W_total = mg (2h + d)

b) The change in gravitational potential energy

           U = mg Δy

in the part with accelerated movement

           U₁ = mg h

in the part with uniform movement

            U₂ = mg d

the total potential energy is

           E_total = U₁ + U₂

           E_total = - mg (h + d)

the answer is a. the yacht traveled from the dock north to the 200-meter marker in the bay in less than 5 minutes, giving the passengers several more hours to fish.

im doing peak rn so sry if its wrong

a straight line sloped