James Jumps: Calculating the Fall in Seconds & Meters

In summary, John jumps out of a window to escape the fire, but his fall is slowed by 1/2 of the gravitational force of Earth.
  • #1
Creaver
12
1

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
 
Last edited:
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  • #2
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.)
 
  • #3
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 !
 
  • #4
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.
 
  • #5
But there are no numbers on my graph...
I dont't have the distance traveled during 1 second.
 
  • #6
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.
 
  • #7
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 ?
 
  • #8
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:
[itex](T-1)^2 = T^2/2[/itex]
Do the algebra.
 
  • #9
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
 
  • Like
Likes PietKuip
  • #14
Poor John (or James?).
May he rest in peace.
 

1. How is the fall time and distance calculated in James Jumps?

The fall time and distance in James Jumps are calculated using the formula t = sqrt(2h/g), where t is the fall time in seconds, h is the height in meters, and g is the acceleration due to gravity (9.8 m/s^2).

2. What is the significance of calculating the fall time and distance in James Jumps?

The calculation of fall time and distance in James Jumps provides important information about the speed and impact of the fall, which can be used to assess the safety of a jump and the potential for injury.

3. Are there any factors that can affect the accuracy of the fall time and distance calculation in James Jumps?

Yes, there are several factors that can affect the accuracy of the calculation, such as air resistance, variations in the acceleration due to gravity, and the landing surface. These factors should be considered when interpreting the results.

4. Can the fall time and distance calculation in James Jumps be used for any type of jump?

The calculation is specifically designed for free-fall jumps, where the height of the jump is known. It may not be accurate for jumps that involve complex movements or non-free-fall elements.

5. Is there a limit to the height that can be used in the fall time and distance calculation in James Jumps?

Technically, there is no limit to the height that can be used in the calculation. However, for safety reasons, it is recommended to stay within reasonable heights and to always follow proper safety protocols when performing jumps.

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