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Find the time of its fall

  • Thread starter whaler
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  • #1
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I have been having some problems with this question:

An object falls from a height h from rest. If it travels a fraction of the total height of 0.5665 in the last 1.00 s, find the time of its fall.

I started by saying h=(0.5)(g)(T^2)
Then I said that 1-.5665h=(0.5)(g)(T-1)^2

I have no idea if this is even remotly right. Please help.
 

Answers and Replies

  • #2
Kurdt
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You have the correct equation. All you need to do is set T=1 and you know that the distance travelled is 0.5665h. Then you can solve for the height put it back in the original equation and solve for total time.
 
  • #3
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so that it states

h=(.5)(9.81)(1)?????
 
  • #4
Kurdt
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so that it states

h=(.5)(9.81)(1)?????
you would have [tex] 0.5665 h =\frac{1}{2}g[/tex]

Now rearrange for h.

When you find h put it back into the original equation and solve for t.
 
  • #5
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i got an h=8.66
Then i put that in and got a value
t=1.33s

That is not the right answer though.....where am i going wrong???
 
  • #6
Kurdt
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i got an h=8.66
Then i put that in and got a value
t=1.33s

That is not the right answer though.....where am i going wrong???
No worries its my fault. Give me a minute (doing many things at once) and I'll correct my mistake.
 
  • #7
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know the feeling....not a problem
 
  • #8
Kurdt
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Ok so the equation that describes the motion is:

[tex] s=h-\frac{1}{2}gt^2 [/tex]

Now the question tells you that in the final second the particle falls [itex]s=0.5665h [/itex] therefore,

[tex] 0.5665h = h-v_t-\frac{1}{2}g [/tex]

where vt is the velocity at t-1 seconds. You can work out this velocity and should get a quadratic in h.
 
  • #9
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so how would I find the vt though?
would i solve the first equation for h and then plug that into the second?
 
  • #10
Kurdt
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What about other kinematic equations? Do you know any more?
 
  • #11
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my prof does not teach? have not be given any others yet
 
  • #13
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Thank You For All The Help!!!!!
 

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