Calculating Free Fall Distance from Time Interval Using Kinematic Equations

[astenroo]

Another free fall topic :)

Here I don't really know where to start

Homework Statement

A body in free fall covers 64% of the total distance fallen in the last second. From what height did it fall?

Homework Equations

y= y1 + y2, where y1=0,36y and y2=0,64y
y=.5at², where a=g=9.8 m/s²

and this is where I get stuck. In a tv-diagram it would be a straight line from origo, and the area under the line gives the distance fallen. And I'm still stuck. Utilizing the y=.5gt² would give me a parabola, and the area under the curve would give me the velocity with which it hits the ground, but I'm not sure whether that would help me much either.

The Attempt at a Solution

 

[CompuChip]

Suppose that the total distance fallen is h, and that t_f is the total falling time.
Then you can fill in your formula for two time instances, namely t_f and (t_f - 1), for example:
h = (1/2) * 9.81 * t_f
(where * stands for multiplication) and another one for t = t_f - 1.

This will give you two formulas with two unknowns (h and t_f) which you should be able to solve for.

 

[astenroo]

Suppose that the total distance fallen is h, and that t_f is the total falling time.
Then you can fill in your formula for two time instances, namely t_f and (t_f - 1), for example:
h = (1/2) * 9.81 * t_f
(where * stands for multiplication) and another one for t = t_f - 1.

This will give you two formulas with two unknowns (h and t_f) which you should be able to solve for.

Ok, so

(1) h=1/2gt_f² and
(2) 0,36h=1/2g(t_f -1)² => h=13,61t_f² - 27,25t_f + 13,61 substitution into (1) which gives
8,71t_f² -27,25t_f + 13,61 => t_f = 2,50s or 0,624s (can be ruled out since it doesn't make any sense)

t_f = 2,50s gives h=1/2g*2,50²=30,625m. I wonder if I made this one too complicated...

 

[CompuChip]

Looks all right to me, except that I cannot clearly make out if you actually used (t_f - 1)² = t_f² - 2 t_f + 1
rather than
(t_f - 1)² = t_f² - 1

Indeed the time interval < 1 second can be ruled out on physical grounds here.

I don't know if this is too complicated, it is the way that I would have done it though.

 

[astenroo]

Looks all right to me, except that I cannot clearly make out if you actually used (t_f - 1)² = t_f² - 2 t_f + 1
rather than
(t_f - 1)² = t_f² - 1

Indeed the time interval < 1 second can be ruled out on physical grounds here.

I don't know if this is too complicated, it is the way that I would have done it though.

I used (t_f - 1)² = t_f² - 2 t_f + 1, which should be the correct way.

Thanks for the help
-alex