Category: Books

The Fall of a Superhero

Recently I have been reading through a newer edition of Young & Freedman’s University Physics, the main course textbook when I was a physics undergrad. Chapter 2 contains the following basic problem concerning 1-dimensional motion with constant acceleration. Since the book does not contain an answer or solution I decided to post a solution here. I highly recommend this book for undergraduate physics students and anyone who wants an excellent overview of the subject.

Question:

The Green Lantern is standing at the top of a tall building with height h and steps off, falling freely from rest to the ground below. He falls half of the total distance in the last 1.0 seconds of his fall. What is the height of the building?

Tama66 / Pixabay

Solution:

We can start by making a sketch of the problem. I personally find it preferable to re-formulate the problem into a very general timeline, with x the displacement from the starting position, as shown below. We know that he starts with zero velocity at time zero. Halfway down the building he has travelled h/2 metres but we do not yet know how long it took or his velocity at the halfway point. Finally, at the end point we know that the time is equal to the time at the halfway point plus one second, the total displacement is h. We do not yet know the final velocity.  The only force acting on the falling object is gravity so we have constant accleration of g = 9.8 metres per second squared.

 

First half of fall:

We have the following equation for motion with constant acceleration (see the text book for details):

v_{1}^{2} = v_{0}^{2} + 2a(x_{1} - x_{0})

Substituting the known values for v_{0} , a , x_{0} and x_{1}:

v_{1}^{2} = 0 + 2g(\frac{h}{2} - 0) = gh

v_{1} = \sqrt{gh}

Second half of fall:

Now for the second half of the fall: Another equation of motion derived in the text book is

x_{2} = x_{1} + v_{1}(t_{2} - t_{1}) + \frac{1}{2} a (t_{2} - t_{1})^{2}

Substituting known values gives us:

h = \frac{h}{2} + v_{1}(1) + \frac{1}{2} g (1) 

Subtracting h/2 from both sides and tidying up:

\frac{h}{2} = v_{1} + \frac{g}{2}

Combining by substitution:

We can then substitute our expression for v_{1} :

\frac{h}{2} = \sqrt{gh} + \frac{g}{2}  

h = 2\sqrt{gh} + g 

We now have an equation with only one unknown, for h (remember that the acceleration due to gravity, g, is known). This is almost in quadratic form if we recognize that it is equal to

h = 2 \sqrt{g} \sqrt{h} + g 

If we substitute \alpha = \sqrt{h}

then we have a quadratic equation be rearranging so that

\alpha ^{2} - 2 \sqrt{g} \alpha - g = 0

and we can solve for \alpha using the well-known equation:

\alpha = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}

where a=1, b=-2\sqrt{g} and c=-1 .

\alpha = \frac{2\sqrt{g} \pm \sqrt{4g + 4g}}{2} = \frac{2\sqrt{g} \pm \sqrt{8g}}{2} = \frac{2\sqrt{g} \pm \sqrt{4}\sqrt{2}\sqrt{g}}{2} = \frac{2\sqrt{g} \pm 2\sqrt{2}\sqrt{g}}{2} = \sqrt{g} \pm \sqrt{2g}

This means that

\sqrt{h} = \alpha = \sqrt{g} \pm \sqrt{2g} = \sqrt{g}( 1 \pm \sqrt{2} )

The quadratic equation gives us two possibilities for the height of the building. The trick here is to recognize that, not only is it not possible for  the height of a building to be negative, but it is also not possible for the square root of the height of a building to be negative! So the only possible value for h ends up being

The quadratic equation gives us two possibilities for the total displacement h. The trick here is to recognize that, since it wouldn’t make sense for the final displacement to be negative, it also wouldn’t make sense for the square root of the final displacement to be negative. We clearly defined the acceleration to be in the positive direction and since we are starting from zero velocity we cannot end up with negative displacement! So the only valid value for h ends up being

h = g(1  + \sqrt{2} )^{2} = g( 1 + 2\sqrt{2} + 2) = g(3 + 2\sqrt{2}) = 9.8(3 + 2\sqrt{2}) = 57.1 m

 




Still no viable 13.3 inch eReader…. hurry up!

Netronix and, purportedly also Onyx International, are developing 13.3 inch screen eReaders using eInk’s Mobius screen, as rivals to Sony’s expensive, PDF-only DPT-S1 Digital Paper. A new video shows the Netronix device prototype being displayed at a recent trade show:

Unfortunately, it seems that without a major company to put up the money to manufacture, market, distribute and sell it the device won’t become available to consumers. That is a real shame! Many people, myself included, would love to have such a device for taking notes or reading scientific papers, magazines and textbooks that don’t display well on, say, a Kindle.

However, I think there is one weakness of the device. Although I haven’t had the privilege of demoing one, the problem also exists on my Onyx Boox M96. Writing with the stylus is not actually that good! Because there is a delay in the line/text appearing as you write it is very hard to write accurately. This seems to make writing small text very difficult since the required stylus strokes are shorter and writing is quicker. The lady in the above video seems to avoid writing small, maybe subconsciously since she already knows that it doesn’t really work well! The manufacturers should really try to improve this aspect of their devices. Perhaps a faster refresh mode while writing would help. Accurately writing with the stylus is essential when notating articles or books, perhaps between the lines of text.

Maybe the resolution of my M96 screen is too low, so that the drawing line cannot be made thinner for more precise drawing and writing. The Mobius screen is supposed to be higher resolution, so that could fix it.

Another crucial problem with eReaders for textbooks is the navigation inside the book, being able to quickly jump between different parts of the book and back again. Think of working on some problems in a Maths textbook and trying to jump to the back for the solutions every few minutes! The software on the device is therefore of utmost importance.

Having stated these problems, I would probably buy one immediately anyway. Please make it available soon!

Update: It looks like Onyx were serious about their 13.3 inch device! It is expected for release in Spring 2016!




Course of Theoretical Physics available online

A collection of classic theoretical physics books, the Course of Theoretical Physics by Landau, Lifshitz, Pitaevskii and Berestetskii (English edition), is available online courtesy of the Internet archive.  It seems that the last two books in the 10 book series are not available, perhaps because their later publication dates meaning they have a longer copyright term. For a discussion on the copyright details of the books and the original source for my links see this Reddit thread. The Russian versions can apparently be found here although I have not verified this.