Falling object Kinematics motion

AI Thread Summary
A falling object travels one-fourth of its total distance during the last second of its fall, prompting a discussion on calculating the height from which it was dropped. The initial assumption is that the initial velocity (vi) is zero, but this is debated since the object has one second remaining in its fall. Participants suggest using kinematic equations to establish two equations for the height (H) and the time (T) to solve for both unknowns. The discussion emphasizes the need to clarify the initial conditions and the relationship between the distances traveled at different intervals. The conversation concludes with a focus on ensuring the correct application of kinematic principles to derive the solution.
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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
 
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Are you sure that initial velocity is zero?
 
mishek said:
Are you sure that initial velocity is zero?

It was not specified so I went in with the assumption vi = 0.
 
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.
 
mishek said:
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?
 
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
 

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mishek said:
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
 
Hello, please check attached photo. Does that make sense?
 

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mishek said:
Hello, please check attached photo. Does that make sense?

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