WebAssign Printed Access Card for Serway/Vuille's College Physics, 11th Edition, Multi-Term
WebAssign Printed Access Card for Serway/Vuille's College Physics, 11th Edition, Multi-Term
11th Edition
ISBN: 9781337763486
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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Chapter 5, Problem 49P

An 80.0-kg skydiver jumps out of a balloon at an altitude of 1.00 × 103 m and opens the parachute at an altitude of 200.0 m. (a) Assuming that the total retarding force on the diver is constant at 50.0 N with the parachute closed and constant at 3.60 × 103 N with the parachute open, what is the speed of the diver when he lands on the ground? (b) Do you think the skydiver will get hurt? Explain. (c) At what height should the parachute be opened so that the final speed of the skydiver when he hits the ground is 5.00 m/s? (d) How realistic is the assumption that the total retarding force is constant? Explain.

(a)

Expert Solution
Check Mark
To determine
The speed of the diver when he lands on the ground.

Answer to Problem 49P

The speed of the diver when he lands on the ground is 24.5m/s .

Explanation of Solution

Given Info:

The mass of the sky diver is 80.0kg .

The height of the balloon from which the sky diver jumps is 1.0×103m .

The parachute opens at an altitude of 200.0m .

The total retarding force when the parachute is open is 50.0N .

The total retarding force when the parachute is closed is 3.60×103N .

According to the Work-Energy theorem in the entire trip,

Wnc=(KE+PEg)f(KE+PEg)i

Since, the retarding force is opposite to the motion of the diver and the final potential energy and initial kinetic energy of the diver is zero. Because the diver is starting from rest and the final position is to the ground.

(F1cos180°)d1+(F2cos180°)d2=(12mvf2+0)(0+mghi)

  • F1 is the total retarding force when the parachute is open
  • F2 is the total retarding force when the parachute is closed
  • d1 is the distance that the diver travels with closed parachute
  • d2 is the height from which the parachute is open
  • m is the mass of diver
  • hi is the initial height of the diver
  • vf is the speed of the diver when he lands on the ground

On re-arranging,

Formula to calculate speed of the diver when he lands on the ground is,

vf=2(ghiF1d1+F2d2m)

Substitute 9.8m/s2 for g , 1.0×103m for hi , 50.0N for F1 , 8000m for d1 , 3.60×103N for F2 , 200.0m for d2 and 80.0kg for m to find the speed of the diver,

vf=2[(9.8m/s2)(1.0×103m)(50.0N)(8000m)+(3.60×103N)(200.0m)80.0kg]=24.5m/s

Thus, the speed of the diver when he lands on the ground is 24.5m/s .

Conclusion:

The speed of the diver when he lands on the ground is 24.5m/s .

(b)

Expert Solution
Check Mark
To determine
Whether the skydiver will get hurt.

Answer to Problem 49P

The landing speed 24.5m/s is enough to get hurt.

Explanation of Solution

Given Info:

The speed of the diver when he lands on the ground is 24.5m/s .

The speed of the skydiver is more than enough to get hurt. At the landing speed of 24.5m/s , he can’t balance in his feet. He will fall down and will get hurt.

Conclusion:

The landing speed 24.5m/s is enough to get hurt.

(c)

Expert Solution
Check Mark
To determine
The height with which the parachute should be opened if he needs a landing speed of 5.0m/s

Answer to Problem 49P

The height with which the parachute should be opened if he needs a landing speed of 5.0m/s is 206m .

Explanation of Solution

Given Info:

The mass of the sky diver is 80.0kg .

The height of the balloon from which the sky diver jumps is 1.0×103m .

The total retarding force when the parachute is open is 50.0N .

The total retarding force when the parachute is closed is 3.60×103N .

The speed of the diver at the ground is 5.0m/s .

According to the Work-Energy theorem in the entire trip,

Wnc=(KE+PEg)f(KE+PEg)i

Consider the height from which the parachute is a variable.

d1=1000md2

  • d1 is the distance that the diver travels with closed parachute

Since, the retarding force is opposite to the motion of the diver and the final potential energy and initial kinetic energy of the diver is zero. Because the diver is starting from rest and the final position is to the ground.

(F1cos180°)(1000md2)+(F2cos180°)d2=(12mvf2+0)(0+mghi)

  • F1 is the total retarding force when the parachute is open
  • F2 is the total retarding force when the parachute is closed
  • d2 is the height from which the parachute is open
  • m is the mass of diver
  • hi is the initial height of the diver
  • vf is the speed of the diver when he lands on the ground

On re-arranging,

Formula to calculate the height with which the parachute should be opened is,

d2=(mg)hi(1000m)f112mvf2F2F1

Substitute 9.8m/s2 for g , 1.0×103m for hi , 50.0N for F1 , 3.60×103N for F2 , 5.0m/s for vf and 80.0kg for m to find the speed of the diver,

d2=[(80.0kg)(9.8m/s2)](1.0×103m)(50.0N)12(80.0kg)(5.0m/s)23.60×103N50.0N=206m

Thus, the height with which the parachute should be opened if he needs a landing speed of 5.0m/s is 206m .

Conclusion:

The height with which the parachute should be opened if he needs a landing speed of 5.0m/s is 206m .

(d)

Expert Solution
Check Mark
To determine
how realistic is the assumption that the total retarding force is a constant.

Answer to Problem 49P

The retarding force will be depending on the speed of the diver in reality.

Explanation of Solution

In the actual case of sky diving, the air drag will be in accordance with the speed of the skydiver.

The weight of the skydiver is about (80.0kg)(9.8m/s2)N

Before the parachute open, the air drag will be less than the weight of the diver.

The air drag will be almost equal to the weight of the diver when he opens the chute and again before he touches down.

Conclusion:

The retarding force will be depending on the speed of the diver in reality.

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Chapter 5 Solutions

WebAssign Printed Access Card for Serway/Vuille's College Physics, 11th Edition, Multi-Term

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