Calculating Skydiver's Landing Speed with Constant Retarding Forces

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Homework Help Overview

The problem involves calculating the landing speed of a skydiver who experiences different retarding forces during free fall and parachute deployment. The scenario includes an initial jump from 1000m, with the parachute opening at 200m, and requires consideration of energy principles such as work done, kinetic energy, and potential energy.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of energy equations, including work done and potential energy, to find the landing speed. There is a focus on the distinction between forces and potential energy, as well as the calculation of work done over different distances.

Discussion Status

Some participants have provided calculations and attempted to clarify the concepts involved. There is an ongoing exploration of the correct values for potential energy and work done, with some corrections noted in the calculations presented.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the information they can share or the methods they can use. There is a focus on ensuring that the calculations align with the principles of physics being discussed.

Paymemoney
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Homework Statement


An 80.0kg skydiver jumps out of balloon at an altitude of 1000m and opens the parachute at an altitude of 200m. a) Assume that the total retarding force on the diver is constant at 50.0N with the parachute closed and constant at 3600N with the parachute open, what is the speed of the diver when he lands on the ground?


Homework Equations


Work Done equation
Kinetic Energy equation
Potential energy equation


The Attempt at a Solution



i have tried to solve this question but my answers does not seem correct and logically not possible. So i need some help on how to start this question.

P.S
 
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Since force is constant and motion is one dimensional, our best friend in this case is the F*d = W, mgh = PE, and 0.5mv^2 = KE equations. There is "positive" energy from the used potential energy in the fall and there is "negative" energy (force is acting opposite to direction of motion) from 800m of constant 50N force and then 200 m of constant 3600N force thereafter. Sum up the "positive" and "negative" energies and you will be able to solve for velocity with 0.5mv^2 = KE (all of the energy is converted to KE).
 
so is the 3600N and 50N Potential Energy or is it the total force exerted within that distance?
 
Force. The potential energy incolves the 1000 m of height.
 
this is what i have done:

PE = mgh
PE = 80 * 9.8 * 1000
PE = 1784J

Now to find the W done for 800m and 200m

W = 50 * 800
W = 4000J negative because it is opposite to the downward direction

W = 3600 * 200
W = -720000J negative because it is opposite to the downward direction

So if i make Total Force=KE then

-4000 + -720000 + 1784 = 0.5* 80 * v^2

722216 = 40 * v^2

v=134.37m/s
 
Thats quite a big error you have in calculating for potential energy there. The concept is right. The calculations are wrong.
 
Paymemoney said:
this is what i have done:

PE = mgh
PE = 80 * 9.8 * 1000
PE = 784000Joules

Now to find the W done for 800m and 200m

W = 50 * 800
W = -40000J negative because it is opposite to the downward direction

W = 3600 * 200
W = -720000J negative because it is opposite to the downward direction

So if i make Total Force=KE then

-40000 + -720000 + 784000 = 0.5* 80 * v^2

24000= 40 * v^2

v=24.5m/s

fixed i get the correct answer :) thanks.
 

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