Conceptual Physics Question

[MedPR]

Ok so this was a question we did in my physics 1 class today. I'm retaking it because I got a C- the first time and because I need the help on MCAT. We worked out the answer some long ass round about way that just won't work for MCAT, but I've seen this question in the prep books and I guess I don't understand it as well as I thought I did.

I'm asking here not because I need the answer for class (we did it as a class today), but because I want to know how you would approach this on the MCAT.

You drop a ball from a 15 foot high building and at the same time another person throws a ball directly upward from the ground below. The initial velocity of that ball is 25m/s. At what time do the balls come into contact with each other?

Edit: Sorry, I should've said the answer is t=0.6seconds.

I'll post the work if anyone wants to see it.

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[milski]

10t^2/2=5-25t-10t^2/2
10t^2+25t-5=0
2t^2+5t-1=0
t=(-5 +/- sqrt(25+8))/2
t=(sqrt(33)-5)/2
t=slight less than one/2
t=0.4

That's with some rounding - 15 ft = 5m. And I have not checked it for math errors.

To be honest, I don't see many options for shortcuts here.

And to answer you more directly - that's what I would write on some scratch paper to solve it. We do get some sort of paper for calculations like this, right?

And one mistake, so I should probably slow down.
t = (sqrt(33)-5)/4
t=0.2

 

[mcloaf]

Hmm, that's an interesting question. I didn't know so I rustled up this (http://www.physicsforums.com/showthread.php?t=5963). But I have no idea how, or if, you could solve this in a manner amenable to the MCAT. Sorry. :/

 

[MedPR]

10t^2/2=5-25t-10t^2/2
10t^2+25t-5=0
2t^2+5t-1=0
t=(-5 +/- sqrt(25+8))/2
t=(sqrt(33)-5)/2
t=slight less than one/2
t=0.4

That's with some rounding - 15 ft = 5m. And I have not checked it for math errors.

To be honest, I don't see many options for shortcuts here.

And to answer you more directly - that's what I would write on some scratch paper to solve it. We do get some sort of paper for calculations like this, right?

And one mistake, so I should probably slow down.
t = (sqrt(33)-5)/4
t=0.2

Yea there was quite a bit of work. Setting up equations for both balls then subtracting one equation from the other. For some reason you can set the final velocity of the two balls equal and also set the time in the air equal for each ball.

Hmm, that's an interesting question. I didn't know so I rustled up this (http://www.physicsforums.com/showthread.php?t=5963). But I have no idea how, or if, you could solve this in a manner amenable to the MCAT. Sorry. :/

Thanks, I'll take a look.

There was another thread on here a while ago about a similar EK 1001 question. I think that one was about a ball thrown up and a ball dropped from a two story building. No initial velocity was given for that problem though.

 

[chiddler]

that's because you don't need initial velocity. you have final velocity = 0, right? can use that.

 

[MedPR]

that's because you don't need initial velocity. you have final velocity = 0, right? can use that.

Can you demonstrate?

 

[milski]

Yea there was quite a bit of work. Setting up equations for both balls then subtracting one equation from the other. For some reason you can set the final velocity of the two balls equal and also set the time in the air equal for each ball.

Sometimes doing the work might be the fastest way. At least I have examples where I've wasted time on exams trying to find a simpler/faster way where I could have been long done just taking the brute force approach.

You can set the times to be equal - that should be straight forward, since the balls are thrown at the same time and hit each other at the same time.

The velocities will be equal only in some very specific reference frame and I honestly cannot think of anything very useful if use that reference frame. They certainly don't have to be the same if the velocity of the ground is considered to be 0 (which is the most typical case).

Thanks, I'll take a look.

There was another thread on here a while ago about a similar EK 1001 question. I think that one was about a ball thrown up and a ball dropped from a two story building. No initial velocity was given for that problem though.

But in that case they told us that the bottom ball was thrown hard enough to reach exactly the height from which the top ball was dropped. That gives you things like the fact that the balls will meet at half of the time it takes for the ball to fall from top to ground and that the final speed at the ball dropping the ground is the same as the initial speed the ball thrown up. Not the case with this problem, so you have to do the calculations.

 

[milski]

that's because you don't need initial velocity. you have final velocity = 0, right? can use that.

That was the other problem where it was said that the ball is thrown up hard enough to reach the exact height from which the first ball is dropped.

 

[chiddler]

That was the other problem where it was said that the ball is thrown up hard enough to reach the exact height from which the first ball is dropped.

oh damn you're right.

 

[MedPR]

Well I'm glad that this problem seems to be a non-mcat-shortcut-applicable problem. I understand how it is solved with all the work, but I thought I was missing something we should know for mcat that would make it quicker.

I don't mind doing the work, but I don't want to waste time unnecessarily.

 

[milski]

Well I'm glad that this problem seems to be a non-mcat-shortcut-applicable problem. I understand how it is solved with all the work, but I thought I was missing something we should know for mcat that would make it quicker.

I don't mind doing the work, but I don't want to waste time unnecessarily.

Practice basic algebra - quadratic equations, factoring out and expanding expressions, proficiency in that will really pay off.

The answer is 0.6? Is the height by any chance 15 m instead of 15 ft? Since I'm pretty far off with 0.2.

 

[MedPR]

Practice basic algebra - quadratic equations, factoring out and expanding expressions, proficiency in that will really pay off.

The answer is 0.6? Is the height by any chance 15 m instead of 15 ft? Since I'm pretty far off with 0.2.

Yea, 15m. Not sure why I typed 15 ft. Sorry