Solving a Free-Fall Problem with Height & Time

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SUMMARY

The discussion focuses on solving a free-fall problem involving height and time, specifically determining the total height (h) and time (T) of an object's fall given that it travels 0.57h meters in the last second. The correct approach involves using the equations of motion: h = 0.5gT^2 and the distance fallen in the last second as 0.57h = 0.5gT^2 - 0.5g(T-1)^2. The calculations reveal that the object falls approximately 8.59 meters and takes about 1.325 seconds to reach the ground, correcting the initial assumption that the object fell 0.57h meters in the first second.

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lolrelativity
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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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