Shorts: Vertical Travel Distance of Bouncing Objects

If a ball is dropped from a height of 10 meters, and on each bounce it reaches a maximal height of 0.75 times the previous height, then what is the total distance traveled? If we use h_i as the maximal height reached on bounce i, h_0 as the initial height, and $latex b $ as the factor of the maximal height achieved relative to the previous bounce, we have h_i = b h_{i-1}.

Then, the total distance traveled is:


\begin{aligned}
d &= h_0 + 2 h_1 + 2 h_2 + 2 h_3 + ...
\\
&= h_0 + 2 b h_0 + 2 b h_1 + 2 b h_2 + ...
\\
&= h_0 + 2 b h_0 + 2 b b h_0 + 2 b b b h_0 + ...
\\
&= h_0 + 2 b h_0 + 2 b^2 h_0 + 2 b^3 h_0 + ...
\\
&= -h_0 + 2 (h_0 + b h_0 + b^2 h_0 + b^3 h_0 + ...)
\\
&= -h_0 + 2 h_0 (1 + b + b^2 + b^3 + ...)
\\
&= -h_0 + 2 h_0 \sum_i \left(b^i\right)
\end{aligned}

Using \sum_n \left(x^n\right) = \frac{1}{1-x} if 0 \le x < 1, we get:


d = -h_0 + 2 h_0 \frac{1}{1-b}

Now we plug our values of h_0 = 10 \text{ m} and b = 0.75 into this equation to find d = -10 + 20 \cdot \frac{1}{0.25} = 70 meters. So, even though our ball will bounce on for eternity, it will travel exactly 70 meters!

Relative Velocity (Or: The Velocity-Addition Formula)

With the knowledge that the speed of light is constant and the same for every reference frame, and that no object can travel at the speed of light or faster than the speed of light in any reference frame, we ask ourselves the question: what happens when two spaceships leave from a space station in opposite directions and both reach a constant speed of 0.95c (with c the speed of light) relative to that space station? In the space station’s reference frame, the two ships travel at 2 \cdot 0.95c = 1.9c relative to each other. This doesn’t violate relativity, as neither spaceship is actually going at or faster than the speed of light. But, ignoring the effects of relativity, in the reference frame of either spaceship, the other ship would appear to travel at 1.9c. This would violate relativity!

Two ships and a space station
Two ships and a space station

Applying the equations we found before, we will find out what actually happens.

Continue reading “Relative Velocity (Or: The Velocity-Addition Formula)”

The Kessel Run in Less than Twelve Parsecs

Millennium Falcon
You’ve never heard of the Millennium Falcon?… It’s the ship that made the Kessel Run in less than twelve parsecs. – Han Solo

If you’ve seen Star Wars: A New Hope, you’ll have heard the above quote. People often think Solo refers to his ship’s speed when he says this, but that is not the case. A parsec is a unit of length, not of time! In fact, it is defined as the distance from the sun to an astronomical object such that the parallax angle (the angle between lines from the Earth and the Sun to that object) is one second of arc.

Parallax
One parsec is the distance from the Sun to a nearby astronomical object that has a parallax angle of 1 second of arc.

Using this information, we will calculate the length of a parsec in meters.

Continue reading “The Kessel Run in Less than Twelve Parsecs”