Question 1090435
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A body travels half of its path length in last second of its free fall. Calculate the height from which it falls ?
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<pre>
It is very well known fact from Physics, or from Calculus, or from your Algebra textbook, 
that a free falling body falls the distance (the height) 

H = {{{(g*t^2)/2}}}


where t is the time of free falling counted from the very beginning time moment.


Let us find the time t based on the given condition.

Then your equation is 

{{{(gt^2)/2}}} - {{{(g*(t-1)^2)/2}}} = {{{H/2}}},     or

{{{(gt^2)/2}}} - {{{g*(t-1)^2)/2}}} = {{{(1/2)*((g*t^2)/2)}}}.


Now cancel the factor "g" in both sides. You will get the last equation in the form

{{{t^2/2}}} - {{{(t-1)^2/2}}} = {{{t^2/4}}}    (*)


    Since we excluded "g" from the equation, the solution for "t" does not depend on units we use 
    for the length or the distance (feet or meters).


Simplify the equation (*)

{{{2t^2}}} - {{{2*(t-1)^2}}} = {{{t^2}}},

{{{t^2 - 4t + 2}}} = 0.


The solution for t is  (use the quadratic formula)

{{{t[1,2]}}} = {{{(4 +- sqrt(4^2 - 4*2))/2}}} = {{{2 +- sqrt(2)}}}.


By the meaning of the condition, the value of "t" must be greater than 1 second, so only the root  t = {{{2 + sqrt(2)}}}  makes sense.


Then the height under the question is

H = {{{(g*t^2)/2}}} = {{{(9.81*(2 + sqrt(2))^2)/2}}} = 57.177 meters.


<U>Check</U>.  Notice that {{{(2+sqrt(2))^2}}} = {{{4 + 4*sqrt(2) + 2}}} = {{{6 + 4*sqrt(2)}}}.

        From the other side,  {{{(1+sqrt(2))^2}}} = {{{1 + 2*sqrt(2) + 2}}} = {{{3 + 2*sqrt(2)}}} is exactly half of that.
</pre>

Solved.



If you want to have the answer in feet, convert from meters to feet.