- #1
mateomy
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Reading a step-by-step problem in my Kinematics chapter; the problem states: "An object is dropped travels one-fourth its distance in the last second of the fall. What height was it dropped from?
They initially use the motion equation:
[tex]
x= x_0 + \frac{1}{2}at^2
[/tex]
(sans the v(Initial) because it is assumed it is dropped from rest)
I can follow that they have to at first solve for time "t" but they move on down the line of variable rearrangement and come to a certain spot where
I HAVE ABSOLUTELY NO IDEA how they go from this...
[tex]
\frac{1}{4}t^2 = t^2 - (t - 1s)^2
[/tex]
to this...
[tex]
(t - 1s) = \pm \sqrt{\frac{3}{4}}
[/tex]
I understand that this might be sort of confusing with no real frame of reference within the problem. I was being to lazy to LaTex all of the steps. I can supply more if requested (ugh).
I've been ruminating over this for the last god-knows-how-long. Maybe I am just fried.
Thanks.
They initially use the motion equation:
[tex]
x= x_0 + \frac{1}{2}at^2
[/tex]
(sans the v(Initial) because it is assumed it is dropped from rest)
I can follow that they have to at first solve for time "t" but they move on down the line of variable rearrangement and come to a certain spot where
I HAVE ABSOLUTELY NO IDEA how they go from this...
[tex]
\frac{1}{4}t^2 = t^2 - (t - 1s)^2
[/tex]
to this...
[tex]
(t - 1s) = \pm \sqrt{\frac{3}{4}}
[/tex]
I understand that this might be sort of confusing with no real frame of reference within the problem. I was being to lazy to LaTex all of the steps. I can supply more if requested (ugh).
I've been ruminating over this for the last god-knows-how-long. Maybe I am just fried.
Thanks.