Another kinematics problem solved at the last second

  • #1
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:

Clausius2 said:
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:
 

Answers and Replies

  • #2
51
0
Nice work, who says maths isn't rewarding?
 
  • #3
Gza
437
0
who says maths isn't rewarding?

That would be me. :wink:
 
  • #4
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.
 

Related Threads on Another kinematics problem solved at the last second

  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
4
Views
4K
  • Last Post
Replies
8
Views
2K
  • Last Post
2
Replies
31
Views
3K
Replies
2
Views
6K
  • Last Post
Replies
8
Views
3K
  • Last Post
Replies
1
Views
1K
Replies
1
Views
1K
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
3
Views
1K
Top