Free fall: finding building height if 1/4 height falls in last second

[Corneo]

I'm really bad at physics so please excuse me. But I would like some help on this question. I would like to be able to solve it without plugging in any numbers until the end.

If a person steps off a building of height h, and free falls on the way down to the bottom. What is the height of the building if he falls a distance of h/4 in his last 1 second of fall.

Any ideas?

 

[Doc Al]

What's the relationship between distance fallen and time for a falling object?

 

[Corneo]

Is it$$t=\frac {2 \Delta x}{ v+v_0}$$?

Then all I have to solve is
$$\frac {h}{4} = \frac {1}{2}g \left(\frac {2 \Delta x}{ v+v_0} - 1 \right)^2 + v_0 \left(\frac {2 \Delta x}{ v+v_0}-1 \right)+h$$

Where $$v_0=0, \Delta x = \frac {3h}{4}$$. Am I correct?

 

[Integral]

Your basic problem is:

$$x(t) = - \frac {g t^2} 2 + V_0 t + X_0$$

$$V_0 =0$$
$$X_0 =h$$

so you get an equation of motion as:
$$x(t) = - \frac {g t^2} 2 + h$$

Solve for t when x=0, this will give you an expression, call it T, for the time of the fall in terms of g and h. Now you know that

$$x(T-1) = \frac h 4$$

use this in your equation of motion to find h.

 

[Corneo]

Oh I see, ok thanks a lot.

 

[KnowledgeIsPower]

Alternatively you can use the four basic constant acceleration equations in a simultaneous equation i think.

 

[thermodynamicaldude]

You could also solve this way:

let t = time it takes to fall.

since x = .5gt^2 (one of the general motion equations)

h = .5g(t^2)

.75h = .5g(t - 1)^2 (since time will be t-1 when it has fallen .75h)

just subtract one equation from another to get .25h = .5(t^2) - .5(t - 1)^2...and simplify to further get...

... h/4 = 2g(2t - 1)...multiply by 4 to get:

h = 2g(2t - 1)...and set this back equal to:

.5g(t^2) = 2g(2t - 1)...this ends up becoming t^2 - 8t + 4 = 0...and then solve for t with the quadratic. Now that you know t, you can easily find h.