Question

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?

Answer 1

`H = 273.4 m`

Step-by-step explanation:

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

So we will have

`H = (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

`(3H)/(4) = (1)/(2)g(n-1)^2`

now from above two equations

`(4)/(3) = ((n)/(n-1))^2`

`1.155(n-1) = n`

`n = 7.46 s`

Now we have

`H = (1)/(2)(9.81)(7.46^2)`

`H = 273.4 m`

Answer 2

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.