Object Free fall quadratic equation problem

[Pupil]

Homework Statement

An object falls from height h from rest. If it travels .50h in the last 1.00 s, find (a) the time and (b) the height of its fall. Explain the physically unacceptable solution of the quadratic equation in t that you obtain.

Homework Equations

Pretty much any free fall equation:
Xf-Xi = .5at^2 + v0t
v^2 = v0^2 + 2a(Xf-Xi)
v = at + v0
Et cetera...

The Attempt at a Solution

I've tried virtually everything I can think of, but no matter how I use/combine the equations I always end up with two or more unknown variables that keeps me from finding (a) or (b). Help! :-(

 

[CompuChip]

For the last part of the freefall, you know Xf-Xi, a and t.
For the first part you know a and v0, and you can calculate v.

 

[Pupil]

For the last part of the freefall, you know Xf-Xi, a and t.
For the first part you know a and v0, and you can calculate v.

So for the first part I know a and v0 and am looking for v, and the equation relating the three is v = at + v0. There are two unknowns here, v and t. How can I find v?

 

[Pupil]

And for the last part, I know Xf-Xi, a and t. But I don't really know Xf-Xi because it is a variable -- h. Also, how do I know t?

 

[ideasrule]

You know how much time the object took to fall from 0.5h to the ground; that's given in the question. To figure out how much time it took in total, you just need to calculate the time taken in the first half of the fall. (Hint: use d=v0t + (1/2)at^2; you know d, v0, and a.)

After calculating the time, you can easily find out the object's speed at 0.5h; after calculating speed, how do you find height?

 

[rl.bhat]

If the total time of fall is t, time for first half is t-1. At the end of this time v is v = vo + a(t-1). And h/2 = 1/2*a(t-1)^2...(1)
This velocity becomes initial velocity for the second part of the fall.
For second half, time of fall is 1 second. Hence
h/2 = a(t-1) + 1/2*a...(2).
Equate equation 1 and 2 and solve the quadratic to find t. Use this t to find h.

 

[Pupil]

Aha! I see. If we use h/2 = ... for the first and second half of the fall, then we can equate them to solve for the unknown! I solved it to get 2 + 2^(1/2) or roughly 3.41s. Thanks so much, guys! I can't believe I couldn't see that.