Free Fall Question: Find Height from Half-Distance Traveled

[Speedking96]

Homework Statement

In the final second of its free fall, an object covers half the height of its total fall. From what height did it fall?

2. The attempt at a solution

I know that the velocity and acceleration at the final second must be enough for the object to cover half the height.

Distance traveled due to acceleration:

= (0.5)(9.81)(1^2) = 4.905 meters.

I understand that whatever the velocity is, it must be enough for the object to travel (h/2) - 4.905 meters. H being the total height.

However, this is where I am having trouble.

 

[lightgrav]

it might make most sense to use fall time as the variable. if the total fall time was 2 seconds, how far would it have gone ... in the first sec, and in the second sec?
... is that half-way ?
ok, what if the total time was 3s? how far in the 1st part (2 s) , compared to the 2nd part (1s)?

 

[Speedking96]

it might make most sense to use fall time as the variable. if the total fall time was 2 seconds, how far would it have gone ... in the first sec, and in the second sec?
... is that half-way ?
ok, what if the total time was 3s? how far in the 1st part (2 s) , compared to the 2nd part (1s)?

So, it is basically trial and error?

 

[hav0c]

no on the contrary you should try it by taking time as a variable.

 

[lightgrav]

you can do it algebraically, if you know what expressions to equate.
If you do a couple of particular examples, you will find out (ie, learn) what those expressions ought to be.
Try it! (I'm not sure you are using the entire distance formula)