BR physics quadratic equation!!

[juligol10]

I was going over the first chapter on BR physics and I notice there is a lot of difficult math without calculators, like square roots and the quadratic equation!

pg.32.. 310m cliff, object leaves top of the cliff at a speed of 35m/s and angle of 50. how far down range is the object when it hits the ground?
I know there has to be an easier way instead of using that freaking equation... any ideas?

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[Will Hunting]

I was going over the first chapter on BR physics and I notice there is a lot of difficult math without calculators, like square roots and the quadratic equation!

pg.32.. 310m cliff, object leaves top of the cliff at a speed of 35m/s and angle of 50. how far down range is the object when it hits the ground?
I know there has to be an easier way instead of using that freaking equation... any ideas?

No, they showed that to show you how NOT to answer that question. Here's how you answer that question. You calculate the initial y velocity component and determine the time until it's 0, max height. You know vof=0 so -voy^2=2*-10H Solve for h then determine time from x=vot +.5(-10)(t^2). Finally, then use their trick for total height drop of 310. 310/5 =62 is about 64 and square root of 64=8 seconds to fall plus original time which is around 3=11s You take 11f*voxff= distance about 250.

Is a real MCAT this involved, perhaps one or 2 questions. However, there just trying to show you how to approach seamingly difficult problems. With practice this problem can be done in 30-45 seconds.

 

[BerkReviewTeach]

I was going over the first chapter on BR physics and I notice there is a lot of difficult math without calculators, like square roots and the quadratic equation!

pg.32.. 310m cliff, object leaves top of the cliff at a speed of 35m/s and angle of 50. how far down range is the object when it hits the ground?
I know there has to be an easier way instead of using that freaking equation... any ideas?

In the upcoming version, they have modified that example to compare/contrast the long-winded school method versus what you need to do on the MCAT. The short cuts are found throughout the chapter, so the attachment assumes you've read them and practiced them. But even without that reference, hopefully the easier route makes sense.

 

[capn jazz]

Yea, it really confused me that they didn't bother to show the non-quadratic way to answer the question. Thanks for the pdf!

 

[juligol10]

ohh thank you! still long.. but more accessible... so when is the next edition coming out? ohh.. and i hope is not always like.. they show you the hard way and not the faster MCAT way...

 

[BerkReviewTeach]

so when is the next edition coming out?

Gawd I don't know with them. I thought it would be out in March, because back in January I was asked to make editting a priority. In fact, it sort of screwed me in lab a bit, because I put more time into the editting than my real job. I assume if it's like their other books, they'll try it out in their class first (I thought that they would have done it this past session, but it doesn't look like it). Then once they get student input, they'll print the final version. I know they are perfectionists, but this is crazy. The book seemed really good when I saw it months ago, so just release it already.

... and i hope is not always like.. they show you the hard way and not the faster MCAT way...

They don't do it often in the new book (I think maybe once every couple chapters), but the few times they do it is to emphasize that you don't need to resort to school methods. I actually think it's a brilliant approach, because in the years I spent teaching, it was always frustrating when people would want to see the long method even though they shouldn't use it. By having it in the book, it appeases the people who have that need. If you don't need to read that, skip to the end and look at the shortcut methods.

 

[ThirdEye]

I'm so glad I found this thread. Being stuck on this part for section 1 has been killing my confidence. If anyone has any info on faster and/or easier methods than what's in the book for upcoming sections.... I'm on my hands and knees. thanks!

 

[meliora27]

In the first example 1.14 how would you set this up to solve. I'm not sure how you can find the range simply knowing the angle and the time it takes to reach the apex. Wouldn't you also need to know the velocity in order to use any of the range equations? Also in example 1.17 it states, "For a 30° launch, the range is a little less than 7 times the maximum height." Where is this reasoning coming from?

 

[BerkReviewTeach]

In the first example 1.14 how would you set this up to solve. I'm not sure how you can find the range simply knowing the angle and the time it takes to reach the apex. Wouldn't you also need to know the velocity in order to use any of the range equations? Also in example 1.17 it states, "For a 30° launch, the range is a little less than 7 times the maximum height." Where is this reasoning coming from?

That's what is what I find so helpful about the new book. The first few examples of range solve the questions as you just described. Then they take the pattern associated with those answers to determine the 4:1 ratio for max height to range when the launch angle is 45 degrees. Once that's established, then the examples start using the shortcut method. It's a great shortcut that if you trust, it can save a great deal of time on these questions.

There's a detailed explanation just before Figure 1-20 and the values for the key angles (30, 45, and 60) are listed in Table 1.4. If you accept that sample calculations as being general enough, then you simply use the shortcut method by comparing values in qestioons to values in that table for various launch angles. For 30 degrees, it gives the R-hmax ratio as 6.7:1, so a 7:1 approximation in the question seems reasonable.

 

[BerkReviewTeach]

I'm so glad I found this thread. Being stuck on this part for lecture 1 has been killing my confidence. If anyone has any info on faster and/or easier methods than what's in the book for upcoming lectures.... I'm on my hands and knees. thanks!

Do you have the new version of the BR book? Are they finally using it in class this summer?

 

[ThirdEye]

Do you have the new version of the BR book? Are they finally using it in class this summer?

I meant to say sections, not lectures. I'm not in the class. I do have the newest versions, I suppose, since I recently got my books... but unfortunately they don't have all of that info you posted in the pdf. Do you have anymore pdfs that fill in missing content?

 

[meliora27]

I got my books w/i the last two weeks and mine were also missing the pdf. That pdf was pretty helpful.

 

[Doodl3s]

Looking at the PDF, would the MCAT even have a physics question that hard? I mean, they know you have only the basic formulas to work with. Second, it looks like BR is basically RETEACHING physics with these tricks of 2s equals ABOUT this much difference, and dividing the range by 4 is the max height... I feel like with BR you'd have to relearn everything!

 

[BerkReviewTeach]

Looking at the PDF, would the MCAT even have a physics question that hard? I mean, they know you have only the basic formulas to work with. Second, it looks like BR is basically RETEACHING physics with these tricks of 2s equals ABOUT this much difference, and dividing the range by 4 is the max height... I feel like with BR you'd have to relearn everything!

While my instinct is to say that such a question is beyond the difficulty level of the MCAT, it would seem that based on feedback here (and a few other places) that every now and then there is a version of the MCAT where such a question would show up. It seems that the actual MACT is a waide range of difficulties. AN dkeep in mind that sometimes a difficult question is quite easy in a multiple choice format. It's a prepare for the worst and best scenario. Learn to solve the simple and challenging questions both.

I'm not sure if reteaching quite describes it, but they definitely present a new way of looking at material. It emphasizes problem solving and concepts in the context of reviewing the material.

And the beauty of a review book is that you only need to review what you don't already know well. You don't have to read everything. If you already know your physics quite well, then skip the review part and just do the practice questions. That, afterall, is the most important part of your review.

 

[lDanny]

I was wondering the same thing.. I have the TPR and the EK books. The TPR seems to show more of a shorter route and omitting the long route. But the TBR has the long route and you have to figure out the short route. Is it like this for all the TBR books? Thanks


also, from the pdf file, (1.15) how did they get the approx. for the height in the turbo method? thanks

 

[capn jazz]

I don't know what you people are talking about. TBR nearly ALWAYS shows the long way AND at least one, if not a few, shortcuts.

 

[ThirdEye]

I don't know what you people are talking about. TBR nearly ALWAYS shows the long way AND at least one, if not a few, shortcuts.

It's just for that one problem. Compare what's in your book to the pdf posted and you'll see what we're talking about.

 

[capn jazz]

I have the books, and I've never seen them tell you to solve a problem in a long, complicated way. 99% of the time, a shortcut is given. Or they tell you how to approach it using intuition. The problem in this thread is a complete anomaly in terms of their usual problems.

 

[bluv1212]

I recently purchased BR Physics and i am also puzzled about the question on page 32. I downloaded the pdf which simplifies the question a lot, but i have a problem with a part of the explanation, as follows:

"From that, we can approximate that the apricot climbs about 40 m from launch to
its apex (based on the approximation that a 1-s climb leads to a height of 5 m, a 2-
s climb leads to a height of 20m, and a 4-s climb leads to a height of 80 m). This
means that at its highest point, the apricot is roughly 350 m above the ground.
We know from our trick on page 21 that it takes 8 s to fall 320 m, so falling 350 m
will take just over 8 s."

where do you get that approximation that every 1 second climb is equal to a 4x height climb?


I can see that this would be helpful, so i'm hoping to get a response soon so that i can move on.

Thanks

 

[PRmed]

I have a question on why my method does not work

I was calculating that if it takes 2.6 seconds to get to the top of the apex, it must take 2.6 seconds to reach the bottom of the apex (the very top of the cliff). So far, that puts us at 5.2 seconds. Then, I would use the y=1/2at^2 but I would plug in 310m for the distance of y (since I had already calculated the time for the entire parabola). That part of the trip would leave me with 7.9 seconds. So 7.9+2.6+2.5=13.1 seconds which makes the answer completely different than 251m. Is there anything that I am missing here, why won't this method work?

 

[ThirdEye]

I have a question on why my method does not work

I was calculating that if it takes 2.6 seconds to get to the top of the apex, it must take 2.6 seconds to reach the bottom of the apex (the very top of the cliff). So far, that puts us at 5.2 seconds. Then, I would use the y=1/2at^2 but I would plug in 310m for the distance of y (since I had already calculated the time for the entire parabola). That part of the trip would leave me with 7.9 seconds. So 7.9+2.6+2.5=13.1 seconds which makes the answer completely different than 251m. Is there anything that I am missing here, why won't this method work?

Because the projectile already has a velocity when it is level with the cliff.

In 2.8 seconds, the projectile climbs about 40m so the total distance it travels down is 350m. If you plug 310m instead of 350m into y=1/2at^2, you are suggesting that the velocity is 0 when it is level with the cliff.

 

[PRmed]

Because the projectile already has a velocity when it is level with the cliff.

In 2.8 seconds, the projectile climbs about 40m so the total distance it travels down is 350m. If you plug 310m instead of 350m into y=1/2at^2, you are suggesting that the velocity is 0 when it is level with the cliff.

Thank you sooo much, its totally starting to make sense now!!

Just a quick question to make sure I get it for any other projectile problem: the value I use for y in y=1/2at^2 will always be the distance once it changes direction (right after the apex)??

 

[ThirdEye]

That's right... otherwise you have to account for the velocity that it already has.

 

[Omega3MD]

Has the PDF been removed from this thread? If so, why? I could really use it since I don't have the new books.

 

[BerkReviewTeach]

Has the PDF been removed from this thread? If so, why? I could really use it since I don't have the new books.

I basically ran out of storage space for PDF files, so I removed the oldest ones. There are other pdfs that show pages from other sections of the book.

 

[Omega3MD]

Cool. Where can I find those?

 

[UIUCstudent]

Cool. Where can I find those?

bump

 

[bonitanolita]

Could you please post a link to the pdfs? They sound really helpful!

 

[flight24]

Could you please post a link to the pdfs? They sound really helpful!

bump

 

[QuiSait]

Bump. I would really like the pdf too.

 

[FutureDoc2]

I could definitely use those pdf files as well! Please share!

 

[Hexon]

No, they showed that to show you how NOT to answer that question. Here's how you answer that question. You calculate the initial y velocity component and determine the time until it's 0, max height. You know vof=0 so -voy^2=2*-10H Solve for h then determine time from x=vot +.5(-10)(t^2). Finally, then use their trick for total height drop of 310. 310/5 =62 is about 64 and square root of 64=8 seconds to fall plus original time which is around 3=11s You take 11f*voxff= distance about 250.

Is a real MCAT this involved, perhaps one or 2 questions. However, there just trying to show you how to approach seamingly difficult problems. With practice this problem can be done in 30-45 seconds.

thanks for the tips
question: why do you need to take the square root of 64?
i'm presuming you're using the s=d/t equation and solving for time: t=d/s where d=310m and s=5m/s as indicated in the question.

btw what is the distance, "11f*voxff"? i understand the 11 is for the 11 seconds, but voxff?

 

[BerkReviewTeach]

thanks for the tips
question: why do you need to take the square root of 64?

It's a BR trick for saving time that was being showcased in that particular example. From the apex of a flight, you can get the drop time using their shortcut. The trick in this question was to first determine the highest point and then get the ascent and descent times using the BR estimation trick. From there, the range is simply vx x ttotal flight.

There are a couple tircks used in that solution (in the newest book) that are really helpful. The question and solution are a little different from the time WillHunting posted.

 

[HerbalPlz]

It's a BR trick for saving time that was being showcased in that particular example. From the apex of a flight, you can get the drop time using their shortcut. The trick in this question was to first determine the highest point and then get the ascent and descent times using the BR estimation trick. From there, the range is simply vx x ttotal flight.

There are a couple tircks used in that solution (in the newest book) that are really helpful. The question and solution are a little different from the time WillHunting posted.

I JUST read the same chapter and question the OP did.

Man, I **** my pants when I saw the page long quadratic equation.

That shortcut technique is pretty cool actually. It's going to take a lot of practice on my end to get used to these techniques though