Solving a Free-Fall Problem with Height & Time

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The discussion revolves around solving a free-fall problem where an object falls a distance h from rest, traveling 0.57h meters in the last second. The initial calculations incorrectly assume the object fell 0.57h meters in the first second, leading to confusion about the total time of fall. The correct approach involves recognizing that the total fall time is T seconds, and the distance fallen in the last second can be expressed using the equations of motion. By applying the equations correctly, one can derive both the total height h and the total time T of the fall. Understanding the distinction between the total fall time and the time for the last second is crucial for accurate calculations.
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I am wondering if I did this right.

Homework Statement



An object falls a distance h from rest. If it travels 0.57h meters in the last 1.00 s,
I need to know the time and the height of its fall.



Homework Equations





The Attempt at a Solution



Here what I have:

Since it traveled .57H meters in the last second, I used y=vt+.5gt^2, and got .57H=4.9 since the time elapsed is 1 second and it came from rest, so the inital velocity is zero. Then I solved for H, getting around 8.59 m. Then with this I set up 8.59=4.9t^2, and solved for T, getting around 1.325 seconds. Was these the right methods?
 
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Since, as you say, the initial velocity is 0, the distance traveled in t seconds is y= .5gt^2. But he "time elapsed" is NOT 1 second. You are told that the object travels 0.57h m in the last second but you don't know how many seconds it had fallen before that. You are assuming the object fell .57h m in the first second.

Suppose it fell for a total of T seconds. Then it fell .57h meters between T-1 and T. In the first T-1 seconds it fell .5g(T-1)^2 and in all T seconds it fell .5gT^2 meters. During that last second, it fell .5gT^2- .5g(T-1)^2= .57h meters. And, of course, you have .5gT^2= h, the total distance fallen.
 
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