Falling object Kinematics motion

  • Thread starter negation
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  • #1
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Homework Statement



A falling object travels one-fourth of its distance during the last second of its fall. From what height was it dropped?

Homework Equations



none

The Attempt at a Solution




yf - yi = vit + 0.5gt^2
0.25x = 0 + 0.5(-9.8ms^-2)(1)^2
x = 19.6m
 

Answers and Replies

  • #2
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Are you sure that initial velocity is zero?
 
  • #3
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Are you sure that initial velocity is zero?

It was not specified so I went in with the assumption vi = 0.
 
  • #4
93
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The problem states that you have a falling object which has one more second to go (last second of it's fall). That means that from that point initial velocity isn't zero.
 
  • #5
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The problem states that you have a falling object which has one more second to go (last second of it's fall). That means that from that point initial velocity isn't zero.

Make sense-should I solve for vf first then?
 
  • #6
93
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you have two unknowns (T & H), you need two equations.

I would make them for two positions shown on a picture (H and 3/4H) by using above mentioned equation:

H=Vo*t + 1/2*g*t^2
 

Attachments

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  • #7
818
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you have two unknowns (T & H), you need two equations.

I would make them for two positions shown on a picture (H and 3/4H) by using above mentioned equation:

H=Vo*t + 1/2*g*t^2

H = vit + 0.5gt^2
1/4H = vit + 0.5gt^2

1/4H = vi(1) + 0.5g(1)^2
vi = 1/4H + 4.9ms^-2
H = (1/4H + 4.9ms^-2)t + 0.5gt^2
 
  • #8
93
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Hello, please check attached photo. Does that make sense?
 

Attachments

  • falling object.png
    falling object.png
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  • #9
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Hello, please check attached photo. Does that make sense?

Why is vi=0 along the 3/4H?
 
  • #10
93
3
Because object starts falling from height H (from rest) where he has initial velocity 0.
 

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