Another kinematics problem solved at the last second

[NoPhysicsGenius]

I am [make that: WAS] having difficulties solving Problem 87 from Chapter 2 of Physics for Scientists and Engineers by Paul A. Tipler, 4th edition. The statement of the problem is as follows:

Ball A is dropped from the top of a building at the same instant that ball B is thrown vertically upward from the ground. When the balls collide, they are moving in opposite directions, and the speed of A is twice the speed of B. At what fraction of the height of the building does the collision occur?​

My incomplete [<Ahem> Make that: COMPLETE] attempt at a solution goes as follows:

We denote the height of the building as h, the position of ball A as y_A, the initial position of ball A as y_A0, and analogously for ball B.

The equation of motion for ball A is the following:
y_A - y_A0 = v_A0t - (0.5)gt²
y_A0 = h
v_A0 = 0
g = 9.81 m/s²
Therefore, y_A = -(0.5)(9.81 m/s²)t² + h
=> y_A = -4.905t² + h

The equation of motion for ball B is the following:
y_B - y_B0 = v_B0t - (0.5)gt²
y_B0 = 0
g = 9.81 m/s²
Therefore, y_B = v_B0t - (0.5)(9.81 m/s²)t²
=> y_B = v_B0t - 4.905t²

The two balls collide when y_A = y_B:

y_A = y_B => -4.905t² + h = v_B0t - 4.905t²
=> h = v_B0t

For ball A:
v_A = v_A0 - gt
v_A0 = 0
Therefore, v_A = - gt => v_A = -9.81t

For ball B:
v_B = v_B0 - gt
Therefore, v_B = v_B0 - gt => v_B = v_B0 - 9.81t

We also know that when y_A = y_B, v_A = -2v_B:

v_A = -2v_B => -9.81t = -2[v_B0 - 9.81t]
=> -9.81t = -2v_B0 + 19.62t
=> 2v_B0 = 29.43t
=> v_B0 = 14.715t

Since h = v_B0t, we have:

h = (14.715t)t = 14.715t²

Unfortunately, I couldn't figure out what to do from this point until I spent the better part of the last hour typing this thread. Then, as I was typing the previous sentence, it hit me:

x_A = -4.905t² + h = -4.905t² + 14.715t²
=> x_A = 9.81t²

Therefore, x_A / h = 9.81t² / (14.715t²)
=> x_A / h = 2/3
=> x_A = 2h/3, which is the answer given in the back of the book.

I've decided to post this thread anyway, partly because someone else might benefit from seeing it, and partly because I spent nearly an hour typing it.

This also reminds me of a post by Clausius2 in a previous thread I had posted:

Maybe the time you have spent writing such [an elaborate] thread you could have re-written your solution of the problem, and surely you would find the error.

I don't think this would have worked the last time, but it certainly worked here!

Oh brother ... Good night, everyone! :zzz:

 

[Gaz031]

Nice work, who says maths isn't rewarding?

 

[Gza]

who says maths isn't rewarding?

That would be me.

 

[Fluorescent]

Hey, I know this threads yonks old, but just wondering, does anyone know where the poster got the very final bit from:

Therefore, xA / h = 9.81t2 / (14.715t2)
=> xA / h = 2/3
=> xA = 2h/3, which is the answer given in the back of the book.

Thanks :)

Edit: Isn't it strange how you see it once you've posted the question haha! Thanks anyway, problem solved.