Why Does an Object Fall 490 Meters in 10 Seconds?

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Homework Help Overview

The discussion revolves around the physics of free fall, specifically examining how far an object falls under the influence of gravity over a period of 10 seconds. Participants are exploring the implications of constant acceleration due to gravity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to apply kinematic equations to determine the distance fallen, with some questioning the appropriate formulas and variables to use. There is discussion about average velocity and the relationship between acceleration and distance.

Discussion Status

The conversation is ongoing, with various approaches being suggested. Some participants have provided formulas and hints, while others are questioning the correctness of the methods and units being used. There is no explicit consensus on the correct approach yet.

Contextual Notes

Participants are navigating issues related to unit conversions and the correct application of kinematic equations. There is a mix of understanding regarding the definitions of distance, velocity, and acceleration in the context of the problem.

tman
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An object falls because of gravity at a rate of -980 cm/sec2 (- sign is indicative of a downward direction) for 10 seconds. If it starts from rest (vo= 0) and its position starts from do= 0, where is it 10 seconds later?

_______________m

I don't know where to start
 
Last edited:
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maybe use d=1/2vt but what do I plug in
 
tman said:
maybe use d=1/2vt but what do I plug in
Whatever is this??

You have constant acceleration, how is distance as a function of time given in that case?
 
I think it is 980m
 
If every second, it is moving 980 cm / second faster, then how fast is it going in 10 seconds?

v = a * t

what is its average velocity during this 10 second peroid? Hint: It started at 0, it ended at a*t. Take the average.

Now distance is easy. It traveled an average velocity for a known time.

d=1/2vt
This formula is wrong.
d = vt

Following the logic above, you should be able to come up with the correct formula for d using a (acceleration) instead of v (veolcity).

I think it is 980m
This is wrong too. How did you get this? Watch your units. You expressed acceleration in cm, but your answer in meters, which is ok to do, but make sure you convert properly.
 
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Just use this formula

[tex]\Delta x = \frac{1}{2}at^2[/tex]

Input the values and solve. -4.9m(100s)

-490 meters from its original position
 
Last edited:

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