# Free fall question

1. Aug 6, 2004

### 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 untill 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?

2. Aug 6, 2004

### Staff: Mentor

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

3. Aug 6, 2004

### 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?

4. Aug 6, 2004

### Integral

Staff Emeritus

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

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

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.

5. Aug 6, 2004

### Corneo

Oh I see, ok thanks a lot.

6. Aug 7, 2004

### KnowledgeIsPower

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

7. Aug 8, 2004

### 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.