Given info: The mass of the person is 64.0 kg, height of the hot air balloon above the ground is 65.0 m, the length of bungee cord is 25.8 m, and spring constant of bungee cord is 81.0 N/m.

The acceleration due to gravity is 9.8 m/s2.

The expression for gravitational potential energy is as follows:

U=mgy

Here,

m is the mass.

g is the acceleration due to gravity.

y is the height.

Substitute 64.0 kg for m and 9.8 m/s2 for g in the above expression.

U=64.0×9.8×y=627.2×y

Conclusion:

Therefore, the gravitational potential energy of the person–Earth system as a function of the person’s variable height y above the ground is 627.2y.

Given info: The mass of the person is 64.0 kg, height of the hot air balloon above the ground is 65.0 m, the length of bungee cord is 25.8 m, and spring constant of bungee cord is 81.0 N/m.

The expression for elastic potential energy of spring is as follows:

Us=12kx2 (1)

k is spring constant.

x is the extension in spring.

The cord will stretch by length x only when the person falls more than the length of the cord. Now, the height of the balloon h should be more than the length of the cord l plus the person’s height y, for safe landing.

The expression for extension in spring is as follows:

x=h−l−y

Here,

h is height of the balloon.

l is the length of cord.

y is person’s height

Substitute 65.0 m for h and 25.8 m for l in the above expression.

x=65.0−25.8−y=39.2−y

Substitute 39.2 m−y for x and 81.0 N/m for k in Equation (1).

Us=12(81.0)(39.2−y)2

Conclusion:

Therefore, the elastic potential energy of the cord as function of y is 12(81.0)(39.2−y)2.

Given info: The mass of the person is 64.0 kg, height of the hot air balloon above the ground is 65.0 m, the length of bungee cord is 25.8 m, and spring constant of bungee cord is 81.0 N/m.

The expression for total potential energy of the person-cord–Earth system is as follows:

UT=U+Us

Substitute 627.2 N×y for U and y2−3.17×103y+6.22×104 for Us in the above expression.

UT=627.2×y+y212(81.0)(39.2−y)2

Conclusion:

Therefore, the total potential energy of the person-cord–Earth system as a function of y is 627.2×y+y212(81.0)(39.2−y)2.

Introduction:

The gravitational potential energy above the surface of the earth is directly proportional to the height of the object.

The elastic potential energy is proportional to the square of displacement.

The total potential energy is the sum of all the potential energies in the system.

Given info: The mass of the person is 64.0 kg, height of the hot air balloon above the ground is 65.0 m, the length of bungee cord is 25.8 m, and spring constant of bungee cord is 81.0 N/m.

From part (a), the expression for gravitational potential energy of person as a function of y is given below:

U=627.2×y.

Table for the above expression is shown below:

The graph of gravitational potential energy with displacement is shown below:

Figure(1)

From part (a), the expression for the elastic potential energy of cord as a function of y is as follows:

Us=12(81.0)(39.2−y)2.

The value of elastic potential energy of the cord remains zero till the person does not fall

equal to the length of cord; therefore, the value of the above equation is zero for y, which is greater than 39.2 m.

Table for the above expression is shown below:

The graph of elastic potential energy with displacement is represented below:

Figure(2)

From part (a), the expression for total potential energy of the person-cord–Earth system as a function of y is as follows:

UT=627.2×y+12(81.0)(39.2−y)2.

When the value of y is greater than 39.2 m, the value of elastic potential energy in the above expression is zero.

Table for the above expression is shown below:

The graph of total potential energy with displacement is represented below:

Figure(3)

Given info: The mass of the person is 64.0 kg, height of the hot air balloon above the ground is 65.0 m, the length of bungee cord is 25.8 m, and spring constant of bungee cord is 81.0 N/m.

The expression for change in total energy is as follows:

ΔU+ΔUs+ΔUk=0

Here,

ΔU is the change in gravitational potential energy.

ΔUs is the change in elastic potential energy.

ΔUk is the change in kinetic energy.

The value of initial kinetic energy is zero, as the person is at rest.

The length of cord is 25.8 m; therefore, the length of cord is subtracted from the height of the balloon.

Substitute −mg(65−y) for ΔU and 12k(39.2−y)2 for ΔUs in the above expression.

−mg(65−y)+12k(39.2−y)2+ΔUk=0

At minimum height above the ground during plunge, the person comes to rest and the change in kinetic energy is zero, as both values of initial and final kinetic energies are zero.

Substitute 0 for ΔUk in the above expression.

−mg(65−y)+12k(39.2−y)2+0=012ky2+(mg−k39.2)y+12k(39.2)2−65mg=0

The expression for the roots of the above quadratic equation is as follows:

h=k39.2−mg±(mg−k39.2)2−4(12k)(12k(39.2)2−65mg)k

Substitute 64.0 kg for m, 9.8 m/s2 for g, and 81.0 N/m for k  in the above expression.     h=(81.0 N/m)39.2−(64.0 kg)(9.8 m/s2)±((81.0 N/m)39.2−(64.0 kg)(9.8 m/s2))2−4(12×81.0 N/m)(12(81.0 N/m)(39.2)2−65(64.0 kg)(9.8 m/s2))(81.0 N/m)=19.8 m

Conclusion:

Therefore, the minimum height of the person above the ground during his plunge is 19.8 m.

Given info: The mass of the person is 64.0 kg, height of the hot air balloon above the ground is 65.0 m, the length of bungee cord is 25.8 m, and the spring constant of bungee cord is 81.0 N/m.

The graph of potential energy shows the equilibrium position at the place where the value of total potential energy is minimum.

The expression for total potential energy is as follows:

UT=627.2×y+12(81.0)(39.2−y)2

Derive the above equation with the height of the person.

dUTdy=627.2−(81.0)(39.2−y)

For the expression of minima, equate the above expression equal to zero.

dUTdy=627.2−(81.0)(39.2−y)=0y=31.46 m

The elevation at the point of equilibrium is 31.46 m.

The person could not stop at the elevation of equilibrium position as he has kinetic energy that does not allow the person to stay at the elevation of equilibrium position. The equilibrium position is unstable.

Conclusion:

Therefore, the potential energy graph shows that in an equilibrium position at an elevation of 31.46 m, the equilibrium position is unstable.

Given info: The mass of the person is 64.0 kg, height of the hot air balloon above the ground is 65.0 m, the length of bungee cord is 25.8 m, and the spring constant of bungee cord is 81.0 N/m.

The expression for change in total energy is as follows:

ΔU+ΔUs+ΔUk=0

Substitute −mg(65−y) for ΔU  and 12k(39.2−y)2 for ΔUs in the above expression.

−mg(65−y)+12k(39.2−y)2+ΔUk=0ΔUk=mg(65−y)−12k(39.2−y)2 (2)

The initial kinetic energy of the jumper is zero; hence, the change in the kinetic energy equals the kinetic energy at that position. Velocity is directly proportional to the square root of the kinetic energy; therefore, for maximum value of kinetic energy, the velocity is maximum.

Differentiate the above expression with respect to y.

dΔUkdy=−mg+k(39.2−y)

Equate dΔUkdy equal to 0 .

dΔUkdy=0

Substitute −mg+k(39.2−y) for dΔUkdy in the above expression.

−mg+k(39.2−y)=0y=39.2−mgk

Substitute 64.0 kg for m, 9.8 m/s2 for g, and 81.0 N/m for k in the above expression.

y=39.2 m−(64.0 kg)(9.8 m/s2)(81.0 N/m)=31.46 m

The height at which the velocity is maximum is 31.46 m.

Substitute 64.0 kg for m, 9.8 m/s2 for g, 81.0 N/m for k, and 31.46 m for y in Equation (2).

ΔUk=(64.0 kg)(9.8 m/s2)(65 m−31.46 m)−12(81.0 N/m)(39.2−31.46 m)2=18610 J

Substitute 12mv2 for ΔUk in the above expression.

12mv2=18610 Jv=2×18610m

Substitute 64.0 kg for m in the above expression.

v=2×18610 J64.0 kg=24.12 m/s

Conclusion:

Therefore, the jumper’s maximum speed is 24.12 m/s.