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:(adsbygoogle = window.adsbygoogle || []).push({});

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 ash, the position of ball A asy, the initial position of ball A as_{A}y, and analogously for ball B._{A0}

The equation of motion for ball A is the following:

yTherefore,_{A}- y_{A0}= v_{A0}t - (0.5)gt^{2}

y_{A0}= h

v_{A0}= 0

g = 9.81 m/s^{2}

y_{A}= -(0.5)(9.81 m/s^{2})t^{2}+ h

=> y_{A}= -4.905t^{2}+ h

The equation of motion for ball B is the following:

yTherefore,_{B}- y_{B0}= v_{B0}t - (0.5)gt^{2}

y_{B0}= 0

g = 9.81 m/s^{2}

y_{B}= v_{B0}t - (0.5)(9.81 m/s^{2})t^{2}

=> y_{B}= v_{B0}t - 4.905t^{2}

The two balls collide wheny:_{A}= y_{B}

y_{A}= y_{B}=> -4.905t^{2}+ h = v_{B0}t - 4.905t^{2}

=> h = v_{B0}t

For ball A:

vTherefore,_{A}= v_{A0}- gt

v_{A0}= 0

v_{A}= - gt => v_{A}= -9.81t

For ball B:

vTherefore,_{B}= v_{B0}- gt

v_{B}= v_{B0}- gt => v_{B}= v_{B0}- 9.81t

We also know that wheny,_{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

Sinceh = v, we have:_{B0}t

h = (14.715t)t = 14.715t^{2}

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:

x_{A}= -4.905t^{2}+ h = -4.905t^{2}+ 14.715t^{2}

=> x_{A}= 9.81t^{2}

Therefore,x, which is the answer given in the back of the book._{A}/ h = 9.81t^{2}/ (14.715t^{2})

=> x_{A}/ h = 2/3

=> x_{A}= 2h/3

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:

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

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

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# Homework Help: Another kinematics problem solved at the last second

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