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?

Question

jamesj · Answer

So suppose the object is dropped from rest at a height of h.

You know that h = (1/2)gT^2, from basic kinematics (yes?). T is the time for the object to reach the zero mark and T is the time taken; g the acceleration due to gravity.

Now, the new information you are given in the problem is that the distance travelled in the last second is h/4. What is that distance travelled in the last second?

Well, at time t = T-1, the object has falled a distance of d1 = (1/2)g(T-1)^2. At time t = T, we know the distance travelled is d2 = (1/2)gT^2 = h. Hence in the last second, the object has travelled a distance of
d2 - d1 = h - (1/2)g(T-1)^2
and you know that is equal to h/4

You will now have to manipulate these equations to eliminate T from them in order to express h purely in terms of g.

jamesj · Answer

...or solve for T in terms of g. Either way!

anonymous · Answer

I'm with you so far... do I eliminate T by using some other form of T?

jamesj · Answer

Well, play around with it and see what you can do. I think I would take this last expression
h - (1/2)g(T-1)^2
and set it equal to h/4 and see what that tells me about T. 

Remember that we can also write h = (1/2)gT^2

anonymous · Answer

Oh, I see! By the time h/4 = h - (1/2)g(T-1)^2 is solved for T, I'll have a new expression to plug in for T.

jamesj · Answer

Yes

anonymous · Answer

Thanks so much. That cleared it right up :)

jamesj · Answer

What answer do you get for h in terms of g?

anonymous · Answer

Actually... I've tried it a few different ways now and something still seems wrong. What you're saying makes sense, but I'm getting just strange answers, like t = sqrt(h/10g)... where did I go wrong?

anonymous · Answer

Or $${3h \over g }=t^2+2t-1$$

jamesj · Answer

Yes, I think you get a quadratic expression. The answer isn't so neat and elegant.

jamesj · Answer

Be careful with signs btw.

anonymous · Answer

Orginally I got $$-{3h \over g} = -t^2-2t+1$$ Should I leave it like that?