Descend Speed After Dropping Glasses: 9.8 m/s

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Homework Help Overview

The problem involves a scenario where a person drops sunglasses from a height of 1400 meters while parachuting, and it takes 15 seconds for the glasses to reach the ground. The original poster is trying to determine their descend speed after dropping the sunglasses, using kinematic equations related to free fall.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate their descend speed by assuming the initial velocity of the glasses is equal to their own descend speed. Some participants question the validity of the problem's parameters, particularly regarding the terminal velocity of the sunglasses and the appropriateness of the given answer key.

Discussion Status

Participants are exploring different interpretations of the problem, with some suggesting that the question may not accurately reflect real-world physics due to the neglect of drag forces. There is a recognition of potential errors in the answer key, but no consensus has been reached on the correct approach or solution.

Contextual Notes

Participants note the constraints of the problem, including the assumption of free fall without accounting for air resistance, which may affect the accuracy of the calculations. There is also mention of the problem being part of a previous physics exam, indicating a potential expectation of simplified conditions.

derpeedoo
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Homework Statement


While parachuting, you dropped your sun glasses at a height of 1400m and it took 15 seconds for the glasses to fall onto the ground. What is your descend speed after you dropped your sun glasses?

Homework Equations


Y = volt + 1/2at^2

The Attempt at a Solution


My thought process: The glasses must have the same velocity as me when I dropped them. Therefore, its initial velocity during the free fall is my descend speed.

1400m = Vo(15s) + .5(-9.8m/s^2)(15s)^2
Vo = 20 m/s

The answer key says my descend speed is 9.8 m/s [/B]
 
Physics news on Phys.org
A book that poses a problem involving parachuting without being cognizant of the fact that the terminal velocity of a pair of sunglasses is far, Far, FAR less than 100 meters per second should be discarded.
 
jbriggs444 said:
A book that poses a problem involving parachuting without being cognizant of the fact that the terminal velocity of a pair of sunglasses is far, Far, FAR less than 100 meters per second should be discarded.
This question was on a previous physics exam. Is there anyway for me to solve this with only knowing the basics of free fall?
 
derpeedoo said:
This question was on a previous physics exam. Is there anyway for me to solve this with only knowing the basics of free fall?
I agree with jbriggs, but in an exam you just have to put up with it. Allowing for drag on the sunglasses would lead to an even larger answer, so I think your method was as intended by the question setter. You got a sign wrong in the first equation but appear to have corrected that, so I also agree with your answer.
 
It's a typo in the answer key. They meant 19.8.
 

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