James Jumps: Calculating the Fall in Seconds & Meters

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James jumps from a building and travels half the height of his fall in one second. The total time of his fall is calculated to be approximately 3.41 seconds, using the formula for distance under constant acceleration due to gravity (g = 9.81 m/s²). The height of the fall is estimated to be around 57.12 meters. The discussion includes attempts to clarify the calculations and the importance of understanding the relationship between time and distance in free fall. The conversation highlights confusion over the calculations and the need for precise algebraic manipulation to arrive at the correct answers.
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Homework Statement


To escape the fire which has just declared itself in the building, James jumps through the window. One second before reaching the mattress, he traveled half of the height of fall.

What is the total time of the fall in seconds ?

What is the total height of the fall in meters ?We shall take g = 9,81 m/s ²

The Attempt at a Solution


2 or 3 secondes ?[/B]
I don't need g... Strange...
my attempt...
1/2gt²

0.5 d= 4.9 (t-1)²
t = 3.4142
d = 57.118
 
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Why do you say 2 or 3 seconds? Do you think he is slowing down or speeding up? In what proportions?
(You do need g for rép2 but not for the first part. And do not delete and repost please.)
 
Strange for rep1 because the answer is x,xxxx...

And no, he accelerates...

One second is the half thus 1 * 2 but it isn't that... and i don't understand why...

somebody has deleted my post, not me !
 
Maybe it was deleted because the heading was so non-descriptive.

I suggest you make a graph of distance versus time and mark the last second before impact.
 
But there are no numbers on my graph...
I dont't have the distance traveled during 1 second.
 
The units on the vertical scale do not matter for how the graph looks like. The first part of the problem only involves the time axis. It is the same on the Moon as on Earth.
 
PietKuip said:
The units on the vertical scale do not matter for how the graph looks like. The first part of the problem only involves the time axis. It is the same on the Moon as on Earth.

I don't understand... One second is the only element in this statement and the answer is decimal...

No Google search is necessary ?
 
Try googling for "John, window"...

Better: call the total time T. The distance is a quadratic function of time, on the Moon, on Earth. So:
(T-1)^2 = T^2/2
Do the algebra.
 
T = 3.41421356

Creaver said:
...
my attempt...
1/2gt²

0.5 d= 4.9 (t-1)²
t = 3.4142
d = 57.118
 
  • #10
and for rep2, d = 57.118 is it correct ?
 
  • #11
No, that is not correct. It is incomplete.
 
  • #12
it is 3:35 AM here, I'm tired.

see you tomorrow and big thank you for your patience !
 
  • #13
d = 1/2 gt² ?
No ?

I have 57.118... meters
 
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  • #14
Poor John (or James?).
May he rest in peace.
 
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