SOLUTION: 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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Question 1090435: A body travels half of its path length in last second of its free fall. Calculate the height from which it falls ?
Found 3 solutions by htmentor, Alan3354, ikleyn:
Answer by htmentor(1343)   (Show Source): You can put this solution on YOUR website!
Let h = the fall height
Using the equation d = vt + 1/2gt^2, if the object falls half its height
in the final second, we can write:
h/2 = v*1 + g/2*1^2 = v + g/2
But the object falls a height of h/2 from rest to reach the speed v:
v^2 = v0^2 + 2gh/2 = gh [v0=0]
So we have two equations in two unknowns. Solve for h:
v^2 = gh
v = 1/2(h-g) -> v^2 = 1/4(h-g)^2
gh = 1/4(h^2 - 2gh + g^2)
h^2 - 6gh + g^2 = 0
The quadratic formula gives:
h = (6g +- sqrt(36g^2 - 4))/2
Using g = 9.81 m/s^2, we get:
h = 58.84 m and h = 0.017 m
The 2nd solution is unrealistic, so we take the 1st.
Ans: h = 58.84 m

Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
A body travels half of its path length in last second of its free fall. Calculate the height from which it falls ?
----------
Using h(t) = 16t^2:
d1 = 16t^2 --- time minus 1 second
d2 = 16(t+1)^2 --- total time falling
--------
d2 = 2d1
---
16(t+1)^2 = 32t^2
t^2 + 2t + 1 = 2t^2
t^2 - 2t - 1 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation (in our case ) has the following solutons:



For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=8 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

Again, the answer is: 2.41421356237309, -0.414213562373095. Here's your graph:

============
t = sqrt(2) + 1 seconds.
h = 16t^2
h = 16*(3 + 2sqrt(2)) = ~ 93.25 feet
=================
Note: The t is exact.
The selection of units (feet, meters, etc) affects the result, as does the approximation of 16t^2.
---
Also, the height cannot be determined unless it's assumed the object impacts after the calculated time.
Answer by ikleyn(52794)   (Show Source): You can put this solution on YOUR website!
.
A body travels half of its path length in last second of its free fall. Calculate the height from which it falls ?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 

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 = 


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 

 -  = ,     or

 -  = .


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

 -  =     (*)


    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 (*)

 -  = ,

 = 0.


The solution for t is  (use the quadratic formula)

 =  = .


By the meaning of the condition, the value of "t" must be greater than 1 second, so only the root  t =   makes sense.


Then the height under the question is

H =  =  = 57.177 meters.


Check.  Notice that  =  = .

        From the other side,   =  =  is exactly half of that.

Solved.


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



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