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Having some conceptual trouble with this problem: "A falling object travels one-fourth of its total distance in the last second of its fall. What height was it dropped from?" Would someone please help me set this one up?

Verified Answer :

MissPhiladelphia

The statement that the object was dropped from a height h (at time t = 0) implies that the initial velocity was zero. The total time t for it to reach the ground is therefore given by one of the constant acceleration equations (s = u.t + 1/2.a.t² with u = 0) 
(s = u.t + 1/2.a.t² with u = 0) 

h = (1/2).g.t² 

where g is the acceleration due to gravity. At time (t - 1) it has fallen three-quarters of the distance h. that is 

3.h/4 = (1/2).g.(t - 1)² 

Eliminating h from these equations produces 

3.h = (3/2).g.t² = 2.g.(t - 1)² 

so g also cancels out and we are left with 

4.(t - 1)² - 3.t² = 0 = t² - 8.t + 4 

which has solutions t = 4 ± 2√3 and hence two heights h = (1/2).g.(4 ± 2√3)², 
where t is in seconds and h in meters when g has units m/s². 

Taking g = 9.8 m/s² gives numerical values for h of (1/2)9.8(28 + 16√3) = 

273.0 m, and (1/2)9.8(28 - 16√3) = 1.41 m. 1.41 m. 

Answer:

\(H = 273.4 m\)

Explanation:

Let the falling object took "n" seconds to reach the ground and it travels H height

So we will have

\(H = \frac{1}{2}gn^2\)

now we know that it covers one fourth of total height in last second

So we can say that it will cover 3H/4 distance in (n-1) seconds

so we will have

\(\frac{3H}{4} = \frac{1}{2}g(n-1)^2\)

now from above two equations

\(\frac{4}{3} = (\frac{n}{n-1})^2\)

\(1.155(n-1) = n\)

\(n = 7.46 s\)

Now we have

\(H = \frac{1}{2}(9.81)(7.46^2)\)

\(H = 273.4 m\)

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