Hi missedinger, that's a totally fair question, which I've addressed in the FAQ: It would be enormously time consuming to also answer all the "General Problems" so I'm limiting coverage only to the regular "Problems", of which there are still more than 1700. If you need help with a "General Problem", my suggestion would be to try and find a regular "Problem" that is similar to the "General Problem", and see if you can apply the same problem solving technique.

Hello, thanks for the question. The lady has two heels, so the weight supported by one heel is half the lady's weight, which works out to $\dfrac{1}{2}mg$.

The only thing that matters is that you're consistent with your coordinate system. This is to say that everything pointing in a particular direction needs the same sign (be it positive or negative doesn't matter, just that they're all the same), and things pointing in the opposite direction have the opposite sign. So, yes, you could make the capsule velocity negative, keeping in mind that this now means the force exerted on the capsule by the astronaut would also be negative in that coordinate system, since it would be directed to the right just the same as the now negative capsule velocity.

Hi cm2hn, we could have used the space capsule final speed in part b), as you suggest, but that's taking a bit of a risk since we calculated the space capsule speed ourselves in part a), and we might have made an error in that calculation. Using the astronaut speed provided by the question is a safer approach, but outside of that, either speed is fine.

When a question doesn't specify an initial speed we can assume it's 0? I'm used to seeing them specify "starting from rest". This is only true with an initial speed of 0.

Hi idan, thank you for the question. When a question doesn't specify an initial speed, and doesn't imply it by saying something like "lifted off the ground" or saying something else which could be interpreted as an initial speed, then you really don't know what the initial speed is. You can't assume it is zero. In this question, the initial speed doesn't actually matter until you get to part e) where it's asking for the final speed. In part e) it finally says "it started from rest". For parts a) through d), the force needed to impart an acceleration of $0.160 \textrm{ g}$ is the same regardless of what speed it's going initially. Only the acceleration is relevant. For example, consider that the force needed to keep the load at rest would be the same as the force needed to make the load rise at a constant speed. The constant speed doesn't matter so far as the force is concerned.

I don't understand part d. Why aren't we solving for t instead, to get an answer of 26.6 seconds? Shouldn't the time elapsed for the friend always be smaller than the time elapsed for us (the observer)?

Hi phamk, thanks for your question. Part d) could be phrased more explicity. The friend is now watching the Earth based observer zoom past, and is taking note of the time on the Earth observer's watch, which is at different locations between the start and end time. This means the friend is not measuring proper time since the starting event (the Earth base observer's watch at the initial time) is not in the same place as the ending event (the Earth based observer's watch at the end time).

The result of each person reporting the same time elapsed on the other person's watch when their own watch shows 20.0 s passed has some symmetry to it. We expect similar results from each perspective since no one perspective is special compared to the other.

Relativity is definitely confusing since it contradicts our slow moving day-to-day experience!
Hope this helps,
Mr. Dychko

Hi fortunado09, sorry about this one. I misspoke in the video. Force is always in Newtons, of course! I'll make a note above the video for other students.

Ah, and yes, "g's" are assumed to be the acceleration of gravity near the surface of Earth, so it can always be substituted with $9.8 \textrm{ m / s}^s$ unless the question explicitly says something else (such as "on Mars" or "10 000 m above the Earth").

Hi fortunado09, N/m is another way of saying $\textrm{kg / s}^2$. Using one or the other is just a matter of personal preference, and I prefer $\textrm{ N/m}$ since I think it better conveys the idea of a spring constant being the number of Newtons of force the spring will exert for each meter that it's compressed. If you expand the Newtons in $\textrm{ N / m}$ to base units, you'll find am "m" cancels, leaving you with your correct alternative units of $\textrm{ kg / s}^s$.

There is no force in the positive x-axis since the shove that made the box initially start moving has already finished. We begin the question at the moment when the shove is finished and the box has an initial speed of $3.5 \textrm {m/s}$, and we want to know how far it will go. If there was no friction it would move forever (that's Newton's first law). However, there is in fact a net force along the x-axis: it's friction since that's the only x-axis force! We calculate how far the box will go given it's initial speed imparted by the shove, and the acceleration (or deceleration if you prefer that term) due to the friction force.

I don't understand, conceptually, how a 2kg box can not move pulling a 5kg box but move it at constant speed pulling a heavier box. Can you please clarify this? Thank you.

Hi Nategrana, thanks for the question. The difference between the scenarios has to do with static friction vs kinetic friction. The main idea to consider in both parts is that the more massive the box on the table, the harder it is to pull it since it will be pressed up upon more strongly by the surface (aka. the Normal force) and thereby experience more friction. Our goal is to find the maximum mass possible before that friction prevents acceleration. Notice that I mention "prevents acceleration", since being at rest or moving at constant speed are both examples of not accelerating, and in either case the friction is fully equivalent in magnitude to the tension resulting from the hanging mass.

The mass of the box on the table is only one part of the issue, and the other part is the coefficient of friction. The higher the coefficient of friction, the more friction force results from a given mass. So, in part a) we're dealing with the box on the table at rest, and this means the coefficient of friction to use is the coefficient of static friction. In part b) since the box is sliding, we use the coefficient of kinetic friction. The coefficient of kinetic friction is always lower than the coefficient of static friction. This means that for the force of friction to equal the tension force, the case b) with a smaller coefficient of friction calls for greater mass than compared to the case of part a) where a higher coefficient of static friction means a lower mass is needed to equal that same tension force.

I think there is an error in the final exponents. I have all the same values as you do for the density of the proton, however my final answer keeps coming out to 3.9E14 kg. I am getting a value of 2.055E-4 for the volume of the baseball.

Thank you so much for spotting this lsugden, you're quite correct that I made an error plugging numbers in for the final calculation. I have updated the quick answer.

Thanks again, and best wishes with your studies,
Mr. Dychko

Why do we do scalar addition and not vector addition here? Is it because when we think of potential, we think of a magnitude of available potential occurring as a sort of pool versus, for example, in electric fields, which have an amount and direction its pushing or pulling a charge with a strength?

Well, scalar addition is used here because we're adding potential energies, and potential energy doesn't have direction. I think it might be helpful to consider what "potential" is, in order to make sense of the fact that it's a scalar. The first thing I always like to remind myself of is that when you hear of "potential", it actually is a lazy way of shortening the more complete term, which is "potential difference". Potential difference is the only thing that can ever be measured. With that said, the next obvious question is "difference" compared to what? In the case of a circuit with a battery, the potential difference is between the positive and negative terminals of the battery. In this question with point charges, the difference is between a point infinitely far away, and the given location of the charge. In other words we have a formula $PE = \dfrac{kq_1q_2}{r}$ that tells us the amount of work an external force (like a hand pushing) would need to do in order to move the charge $q_1$ from infinitely far away to it's given location, which requires work since there's a force of repulsion as a result of $q_2$ (assuming the charges are of the same sign). When there are more charges in addition to $q_2$, as there are on the other two corners of the square in this question, we use the formula a total of 3 times and add the results to find the total. The addition is as scalars since we're adding energies, which don't have direction.

Hi antoinec467, It would be enormously time consuming to also answer all the "General Problems" so I'm limiting coverage only to the regular "Problems", of which there are still more than 1700. If you need help with a "General Problem", my suggestion would be to try and find a regular "Problem" that is similar to the "General Problem", and see if you can apply the same problem solving technique.

Why are we looking for d^2 when finding the diagonal distance? The distance of the hypotenuse of a 3,4,5 triangle is not 25 its square root of it which is 5..... A bit confused:(

Hi nategrana, thanks for the question. To determine the 'x' component of the force on the '2Q' charge, we need the 'x' component of the force due to the '4Q' charge diagonally opposite. The force formula has the distance between the '2Q' and '4Q' charge in the denominator, but it is squared. We're looking for the square of the distance since that's the quantity to plug into the force formula.

Hi mcgracia2008, thanks for looking at the working. I think what you're commenting on is the calculator display in which you see $-28^2/2/88$, but this is the same as $-28^2/(2*88)$. I just like the division signs since it's one less character to press instead of using parentheses for the multiplication in the denominator, but that's just a personal preference.

It seems strange, but true, that mass doesn't affect the maximum velocity. Keep in mind that the friction we're speaking about here is normal to the velocity. In other words, the static friction is perpendicular to the direction of motion. We're speaking only about the friction that changes the direction of the car, so this friction has no effect on the speed.

On the other hand, you might be thinking about "rolling friction", or perhaps "air friction", or in other words the frictions that cause things with wheels to slow down and eventually stop. These types of friction are directed opposite to the direction of motion, and so they indeed do slow the car down. These types of friction are not considered in this question. Rolling friction, indeed, would increase with mass.

Hi chaegyunkang, from your other questions it's clear that you're using the Global 7th Edition. I think this particular question was removed from the Global Edition.

Hi choianchoi, I've writen the acceleration as -9.8m/s^2 in the formula, and it's important to stay consistent once you've chosen a coordinate system. That is to say that if down is chosen to be negative, as is often the case, then it has to be considered negative for the entire question. Since we've chosen down as negative, the acceleration due to gravity can never be positive, regardless of where the ball is in it's trajectory.

Hi, as the question asked how far from its base shouldn't the answer be square root (x ^2+y^2)? (displacement between landing point and the jump point)

Hi choianchoi, actually, since the question uses the word base, it's referring to the bottom of the cliff. The bottom of the cliff is the base of the cliff, so the question is asking only for the horizontal component of the displacement from the jump point to the landing point.

## Giancoli 7th Edition, Chapter 6, Problem 3

By missedinger on Sat, 12/02/2017 - 21:05wondering why you stop @ question 71 for solutions, when the text books has 94 questions?

Hi missedinger, that's a totally fair question, which I've addressed in the FAQ: It would be enormously time consuming to also answer all the "General Problems" so I'm limiting coverage only to the regular "Problems", of which there are still more than 1700. If you need help with a "General Problem", my suggestion would be to try and find a regular "Problem" that is similar to the "General Problem", and see if you can apply the same problem solving technique.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 10, Problem 9

By jr59j on Fri, 12/01/2017 - 11:55Why is it 1/2 of m*g

Hello, thanks for the question. The lady has two heels, so the weight supported by one heel is half the lady's weight, which works out to $\dfrac{1}{2}mg$.

Cheers,

Mr. Dychko

## Giancoli 7th Edition, Chapter 7, Problem 19

By cm2hn on Mon, 11/20/2017 - 20:20Why the speed of astronaut is negative? Would I get the same result if I use the capsule as the negative value?

The only thing that matters is that you're consistent with your coordinate system. This is to say that everything pointing in a particular direction needs the same sign (be it positive or negative doesn't matter, just that they're all the same), and things pointing in the opposite direction have the opposite sign. So, yes, you could make the capsule velocity negative, keeping in mind that this now means the force exerted on the capsule by the astronaut would also be negative in that coordinate system, since it would be directed to the right just the same as the now negative capsule velocity.

## Giancoli 7th Edition, Chapter 7, Problem 19

By cm2hn on Mon, 11/20/2017 - 20:06Why average force is based on ma*vfa and not ms*vfs?

Hi cm2hn, we could have used the space capsule final speed in part b), as you suggest, but that's taking a bit of a risk since we calculated the space capsule speed ourselves in part a), and we might have made an error in that calculation. Using the astronaut speed provided by the question is a safer approach, but outside of that, either speed is fine.

## Giancoli 7th Edition, Chapter 6, Problem 25

By idan on Wed, 11/15/2017 - 17:49When a question doesn't specify an initial speed we can assume it's 0? I'm used to seeing them specify "starting from rest". This is only true with an initial speed of 0.

Hi idan, thank you for the question. When a question doesn't specify an initial speed, and doesn't imply it by saying something like "lifted off the ground" or saying something else which could be interpreted as an initial speed, then you really don't know what the initial speed is. You can't assume it is zero. In this question, the initial speed doesn't actually matter until you get to part e) where it's asking for the final speed. In part e) it finally says "it started from rest". For parts a) through d), the force needed to impart an acceleration of $0.160 \textrm{ g}$ is the same regardless of what speed it's going initially. Only the acceleration is relevant. For example, consider that the force needed to keep the load at rest would be the same as the force needed to make the load rise at a constant speed. The constant speed doesn't matter so far as the force is concerned.

Hope this helps,

Mr. Dychko

## Giancoli 6th Edition, Chapter 26, Problem 13

By phamk on Fri, 11/10/2017 - 08:57I don't understand part d. Why aren't we solving for t instead, to get an answer of 26.6 seconds? Shouldn't the time elapsed for the friend always be smaller than the time elapsed for us (the observer)?

Hi phamk, thanks for your question. Part d) could be phrased more explicity. The friend is now watching the Earth based observer zoom past, and is taking note of the time on the Earth observer's watch, which is at different locations between the start and end time. This means the friend is not measuring proper time since the starting event (the Earth base observer's watch at the initial time) is not in the same place as the ending event (the Earth based observer's watch at the end time).

The result of each person reporting the same time elapsed on the other person's watch when their own watch shows 20.0 s passed has some symmetry to it. We expect similar results from each perspective since no one perspective is special compared to the other.

Relativity is definitely confusing since it contradicts our slow moving day-to-day experience!

Hope this helps,

Mr. Dychko

## Giancoli 7th Edition, Chapter 6, Problem 65

By Mr. Dychko on Tue, 11/07/2017 - 10:45Thank you kfcheung1016 and fortunado09! I've corrected the final answer.

## Giancoli 7th Edition, Chapter 6, Problem 20

By fortunado09 on Sun, 11/05/2017 - 17:34The force of the ball on the glove is in joules? The answer is in Newtons, though. What am I missing?

Hi fortunado09, sorry about this one. I misspoke in the video. Force is always in Newtons, of course! I'll make a note above the video for other students.

## Giancoli 7th Edition, Chapter 6, Problem 65

By fortunado09 on Sun, 11/05/2017 - 16:07Caught this one too, where are you?

## Giancoli 7th Edition, Chapter 6, Problem 63

By fortunado09 on Sun, 11/05/2017 - 15:54You’ve got 70km/h in your work, but it’s 80km/h in the problem and your calculator!

Nice catch! I'll mention this error in the final answer. Thanks again for reporting it.

## Giancoli 7th Edition, Chapter 6, Problem 42

By fortunado09 on Sun, 11/05/2017 - 13:11How do you get the units N/m? I’m getting kg/s^2 which is not a thing I’m familiar with.

How do you know to use 9.8m/s^2 per g? Is that just gravity when dealing with g’s?

Ah, and yes, "g's" are assumed to be the acceleration of gravity near the surface of Earth, so it can always be substituted with $9.8 \textrm{ m / s}^s$ unless the question explicitly says something else (such as "on Mars" or "10 000 m above the Earth").

Hi fortunado09, N/m is another way of saying $\textrm{kg / s}^2$. Using one or the other is just a matter of personal preference, and I prefer $\textrm{ N/m}$ since I think it better conveys the idea of a spring constant being the number of Newtons of force the spring will exert for each meter that it's compressed. If you expand the Newtons in $\textrm{ N / m}$ to base units, you'll find am "m" cancels, leaving you with your correct alternative units of $\textrm{ kg / s}^s$.

Cheers,

Mr. Dychko

## Giancoli 7th Edition, Chapter 6, Problem 65

By kfcheung1016 on Mon, 10/30/2017 - 06:31there is a typing error

the ans is 5.4×104 W in (a.

Thank you

5.3×10^4 *

## Giancoli 7th Edition, Chapter 4, Problem 42

By tnreifsnyder on Wed, 10/25/2017 - 21:55If the box is pushed, why is there no force in the positive direction on the x-axis? Also, why is there no net force?

Hi tnreifsnyder,

There is no force in the positive x-axis since the shove that made the box initially start moving has already finished. We begin the question at the moment when the shove is finished and the box has an initial speed of $3.5 \textrm {m/s}$, and we want to know how far it will go. If there was no friction it would move forever (that's Newton's first law). However, there is in fact a net force along the x-axis: it's friction since that's the

onlyx-axis force! We calculate how far the box will go given it's initial speed imparted by the shove, and the acceleration (or deceleration if you prefer that term) due to the friction force.Hope that helps,

Mr. Dychko

## Giancoli 7th Edition, Chapter 4, Problem 47

By nategrana on Mon, 10/02/2017 - 18:31I don't understand, conceptually, how a 2kg box can not move pulling a 5kg box but move it at constant speed pulling a heavier box. Can you please clarify this? Thank you.

Hi Nategrana, thanks for the question. The difference between the scenarios has to do with

static frictionvskinetic friction. The main idea to consider in both parts is that the more massive the box on the table, the harder it is to pull it since it will be pressed up upon more strongly by the surface (aka. the Normal force) and thereby experience more friction. Our goal is to find the maximum mass possible before that friction prevents acceleration. Notice that I mention "prevents acceleration", since being at rest or moving at constant speed are both examples of not accelerating, and in either case the friction is fully equivalent in magnitude to the tension resulting from the hanging mass.The mass of the box on the table is only one part of the issue, and the other part is the coefficient of friction. The higher the coefficient of friction, the more friction force results from a given mass. So, in part a) we're dealing with the box on the table at rest, and this means the coefficient of friction to use is the coefficient of static friction. In part b) since the box is sliding, we use the coefficient of kinetic friction. The coefficient of kinetic friction is always lower than the coefficient of static friction. This means that for the force of friction to equal the tension force, the case b) with a smaller coefficient of friction calls for greater mass than compared to the case of part a) where a higher coefficient of static friction means a lower mass is needed to equal that same tension force.

Hope that helps,

Mr. Dychko

## Giancoli 7th Edition, Chapter 1, Problem 24

By lsugden on Sun, 10/01/2017 - 11:43I think there is an error in the final exponents. I have all the same values as you do for the density of the proton, however my final answer keeps coming out to 3.9E14 kg. I am getting a value of 2.055E-4 for the volume of the baseball.

Thank you so much for spotting this lsugden, you're quite correct that I made an error plugging numbers in for the final calculation. I have updated the quick answer.

Thanks again, and best wishes with your studies,

Mr. Dychko

m^3 for the V_baseball

## Giancoli 6th Edition, Chapter 2, Problem 5

By tahell2000 on Wed, 09/20/2017 - 16:04The answer in top left is labeled m/s . Should be cm/s.

Fixed! Thank you very much for noticing.

## Giancoli 7th Edition, Chapter 17, Problem 22

By nategrana on Thu, 09/14/2017 - 16:08Why do we do scalar addition and not vector addition here? Is it because when we think of potential, we think of a magnitude of available potential occurring as a sort of pool versus, for example, in electric fields, which have an amount and direction its pushing or pulling a charge with a strength?

Well, scalar addition is used here because we're adding potential energies, and potential energy doesn't have direction. I think it might be helpful to consider what "potential" is, in order to make sense of the fact that it's a scalar. The first thing I always like to remind myself of is that when you hear of "potential", it actually is a lazy way of shortening the more complete term, which is "potential difference".

Potential differenceis the only thing that can ever be measured. With that said, the next obvious question is "difference" compared to what? In the case of a circuit with a battery, the potential difference is between the positive and negative terminals of the battery. In this question with point charges, the difference is between a point infinitely far away, and the given location of the charge. In other words we have a formula $PE = \dfrac{kq_1q_2}{r}$ that tells us theamount of workan external force (like a hand pushing) would need to do in order to move the charge $q_1$ from infinitely far away to it's given location, which requires work since there's a force of repulsion as a result of $q_2$ (assuming the charges are of the same sign). When there are more charges in addition to $q_2$, as there are on the other two corners of the square in this question, we use the formula a total of 3 times and add the results to find the total. The addition is as scalars since we're adding energies, which don't have direction.Hope that helps!

## Giancoli 6th Edition, Chapter 2, Problem 52

By antoinec467 on Mon, 09/04/2017 - 10:00Hi how to find the answers for problem 62 and 71?

Hi antoinec467, It would be enormously time consuming to also answer all the "General Problems" so I'm limiting coverage only to the regular "Problems", of which there are still more than 1700. If you need help with a "General Problem", my suggestion would be to try and find a regular "Problem" that is similar to the "General Problem", and see if you can apply the same problem solving technique.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 16, Problem 15

By nategrana on Thu, 08/31/2017 - 12:09Why are we looking for d^2 when finding the diagonal distance? The distance of the hypotenuse of a 3,4,5 triangle is not 25 its square root of it which is 5..... A bit confused:(

Hi nategrana, thanks for the question. To determine the 'x' component of the force on the '2Q' charge, we need the 'x' component of the force due to the '4Q' charge diagonally opposite. The force formula has the distance between the '2Q' and '4Q' charge in the denominator, but it is

squared. We're looking for the square of the distance since that's the quantity to plug into the force formula.Hope that helps,

Mr. Dychko

WOA! Totally missed that. Thank you Mr. Dychko. Great site!

## Giancoli 7th Edition, Chapter 2, Problem 22

By mcgracia2008 on Thu, 08/24/2017 - 18:50I think the result is incorrect. Instead ot multiply 2(88), the tutor divided 2/88.

Hi mcgracia2008, thanks for looking at the working. I think what you're commenting on is the calculator display in which you see $-28^2/2/88$, but this is the same as $-28^2/(2*88)$. I just like the division signs since it's one less character to press instead of using parentheses for the multiplication in the denominator, but that's just a personal preference.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 5, Problem 9

By idan on Thu, 08/03/2017 - 15:20I see how the formula no longer has m in it. However, doesn't friction increase with greater mass which then changes the maximum velocity?

It seems strange, but true, that mass doesn't affect the maximum velocity. Keep in mind that the friction we're speaking about here is

normalto the velocity. In other words, the static friction is perpendicular to the direction of motion. We're speaking only about the friction that changes the direction of the car, so this friction has no effect on the speed.On the other hand, you might be thinking about "rolling friction", or perhaps "air friction", or in other words the frictions that cause things with wheels to slow down and eventually stop. These types of friction are directed opposite to the direction of motion, and so they indeed do slow the car down. These types of friction are not considered in this question. Rolling friction, indeed, would increase with mass.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 11, Problem 21

By chaegyunkang on Tue, 08/01/2017 - 04:47This question is the answer for question 20 not 21.

Hi chaegyunkang, it looks like you're using the Global 7th Edition. Here's the identical solution for problem 20, as you say, in the Global edition: https://www.giancolianswers.com/giancoli-physics-7th-global-edition-solu...

Please navigate for solutions starting with the Global edition: https://www.giancolianswers.com/giancoli-physics-7th-global-edition-chap...

Best wishes,

Mr. Dychko

## Giancoli 7th Edition, Chapter 11, Problem 20

By chaegyunkang on Tue, 08/01/2017 - 04:46This question is the answer for question 19 not 20.

It's listed as #19 in the Global Edition: https://www.giancolianswers.com/giancoli-physics-7th-global-edition-solu..., the navigation for which begins here: https://www.giancolianswers.com/giancoli-physics-7th-global-edition-chap...

Cheers,

Mr. Dychko

## Giancoli 7th Edition, Chapter 11, Problem 19

By chaegyunkang on Tue, 08/01/2017 - 04:43For part a) it is asking for the amplitude of the motion. The video solved for the wrong question and the answer is also wrong.

Hi chaegyunkang, from your other questions it's clear that you're using the Global 7th Edition. I think this particular question was removed from the Global Edition.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 20, Problem 28

By eptrottier on Tue, 07/25/2017 - 12:38where do you get the formula in the blue?

Hi eptrottier, that formula is on page 571, labelled 20-7. It's the force between two parallel current carrying wires.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 3, Problem 30

By choianchoi on Sun, 07/23/2017 - 20:44Hello, why is the acceleration -9.8 m/s^2 when after the ball reaches the top, the acceleration is +9.8m/s^2?

Hi choianchoi, I've writen the acceleration as -9.8m/s^2 in the formula, and it's important to stay consistent once you've chosen a coordinate system. That is to say that if down is chosen to be negative, as is often the case, then it has to be considered negative for the entire question. Since we've chosen down as negative, the acceleration due to gravity can never be positive, regardless of where the ball is in it's trajectory.

Best wishes,

Mr. Dychko

## Giancoli 7th Edition, Chapter 3, Problem 18

By choianchoi on Sun, 07/23/2017 - 02:44Hi, as the question asked how far from its base shouldn't the answer be square root (x ^2+y^2)? (displacement between landing point and the jump point)

Hi choianchoi, actually, since the question uses the word

base, it's referring to thebottomof the cliff. The bottom of the cliff is thebaseof the cliff, so the question is asking only for the horizontal component of the displacement from the jump point to the landing point.Hope that helps,

Mr. Dychko

## Giancoli 7th Edition, Chapter 2, Problem 59

By choianchoi on Sat, 07/22/2017 - 00:57From 30 s to 50 s why is it curved and not straight?

Thanks

## Giancoli 7th Edition, Chapter 4, Problem 8

By chris010727 on Sun, 07/09/2017 - 07:53Why did you set the initial velocity as 0. Isn't it supposed to be 13 instead because the shot will eventually come to a rest.

## Giancoli 7th Edition, Chapter 23, Problem 51

By arielle.lab on Wed, 07/05/2017 - 06:01Hi! How do you know that do is positive?