NoPhysicsGenius
Aug17-04, 11:07 PM
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 yA, the initial position of ball A as yA0, and analogously for ball B.
The equation of motion for ball A is the following:
yA - yA0 = vA0t - (0.5)gt2
yA0 = h
vA0 = 0
g = 9.81 m/s2
Therefore, yA = -(0.5)(9.81 m/s2)t2 + h
=> yA = -4.905t2 + h
The equation of motion for ball B is the following:
yB - yB0 = vB0t - (0.5)gt2
yB0 = 0
g = 9.81 m/s2
Therefore, yB = vB0t - (0.5)(9.81 m/s2)t2
=> yB = vB0t - 4.905t2
The two balls collide when yA = yB:
yA = yB => -4.905t2 + h = vB0t - 4.905t2
=> h = vB0t
For ball A:
vA = vA0 - gt
vA0 = 0
Therefore, vA = - gt => vA = -9.81t
For ball B:
vB = vB0 - gt
Therefore, vB = vB0 - gt => vB = vB0 - 9.81t
We also know that when yA = yB, vA = -2vB:
vA = -2vB => -9.81t = -2[vB0 - 9.81t]
=> -9.81t = -2vB0 + 19.62t
=> 2vB0 = 29.43t
=> vB0 = 14.715t
Since h = vB0t, 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. :grumpy: Then, as I was typing the previous sentence, it hit me:
xA = -4.905t2 + h = -4.905t2 + 14.715t2
=> xA = 9.81t2
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.
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. :cry:
This also reminds me of a post by Clausius2 in a previous thread I had posted (http://www.physicsforums.com/showpost.php?p=286623&postcount=4):
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. :wink:
I don't think this would have worked the last time, but it certainly worked here! :rofl: :biggrin:
Oh brother ... Good night, everyone! :zzz:
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 yA, the initial position of ball A as yA0, and analogously for ball B.
The equation of motion for ball A is the following:
yA - yA0 = vA0t - (0.5)gt2
yA0 = h
vA0 = 0
g = 9.81 m/s2
Therefore, yA = -(0.5)(9.81 m/s2)t2 + h
=> yA = -4.905t2 + h
The equation of motion for ball B is the following:
yB - yB0 = vB0t - (0.5)gt2
yB0 = 0
g = 9.81 m/s2
Therefore, yB = vB0t - (0.5)(9.81 m/s2)t2
=> yB = vB0t - 4.905t2
The two balls collide when yA = yB:
yA = yB => -4.905t2 + h = vB0t - 4.905t2
=> h = vB0t
For ball A:
vA = vA0 - gt
vA0 = 0
Therefore, vA = - gt => vA = -9.81t
For ball B:
vB = vB0 - gt
Therefore, vB = vB0 - gt => vB = vB0 - 9.81t
We also know that when yA = yB, vA = -2vB:
vA = -2vB => -9.81t = -2[vB0 - 9.81t]
=> -9.81t = -2vB0 + 19.62t
=> 2vB0 = 29.43t
=> vB0 = 14.715t
Since h = vB0t, 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. :grumpy: Then, as I was typing the previous sentence, it hit me:
xA = -4.905t2 + h = -4.905t2 + 14.715t2
=> xA = 9.81t2
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.
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. :cry:
This also reminds me of a post by Clausius2 in a previous thread I had posted (http://www.physicsforums.com/showpost.php?p=286623&postcount=4):
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. :wink:
I don't think this would have worked the last time, but it certainly worked here! :rofl: :biggrin:
Oh brother ... Good night, everyone! :zzz: