# Physics for martial arts students

Right after the kickboxing class, I saw one of classmate and friend of mine trying to solve a physics exercise. I studied physics in high school, and I had a exam in college; it was long time ago and even at that time I wasn’t very strong in the matter (and I used to soothe myself (or to fool myself?) quoting Linus Torvalds “While in physics you’re supposed to figure out how the world is made up, in computer science you create the world.” But that’s a different story), but I decided that I could help, after all it didn’t sound too difficult.

### The problem

The problem required finding the final velocity (rounded at one decimal digit), $v_f$ of a ball thrown down from a 40 meters tall tower with a initial velocity, $v_0$, of 12 m/s. The exercise book was providing also the solution: 30.5 m/s.

### The solution

It is very likely that in the very first pages of any physics textbook there is a direct formula to solve the problem with the given data, but, as I mentioned before, I knew little and I remember less; it took around 40 minutes, but we found the solution.

The first step was to write down a formula to determine the final velocity, intuitively it should be given by the sum of the initial velocity and how it changed during the fall; something like the following formula should do the job:

$v_f = v_0 + g\cdot t$     (1)

where $g$ is the gravitational acceleration (9.8 m/s), and $t$ is the time taken for the ball to hit the ground. It looks reasonable, but we didn’t know yet how long it took for the ball to hit the ground, in other words there are two unknown in the above formula: the final velocity, $v_f$ and the duration of the motion $t$.

We didn’t know the time taken $t$ and we didn’t use the information regarding the height of the tower (or in other words the distance covered during the motion); it sounded like a good idea determining $t$ using another formula that uses the distance covered $d$; we knew that the average speed is computed as the distance covered divided by the time taken to cover the same distance:

$v_{avg} = \frac{d}{t}$     (2).

With some simple arithmetic we can obtain the time taken as:

$t = \frac{d}{v_{avg}}$     (3).

Did we know the average velocity? We didn’t! Since the acceleration (we didn’t forget that it is the rate of change of velocity) in a free fall is constant, it looked reasonable to compute the average velocity as the sum of the initial and final velocities and divided it by two:

$v_{avg} = \frac{v_0 + v_f}{2}$     (4).

Using the formula (4) and (3) we obtained a new formula for determining the time taken:

$t = \frac{2\cdot d}{v_0 + v_f}$     (5).

Again two unknowns! But we were lucky the are the same two unknowns seen in (1), this means that we can combine (1) and (5) to get rid of one of them and determine the value of the other; so if we substitute $t$ in (1) with the left side of (5) we obtain:

$v_f = v_0 + g\cdot \frac{2\cdot d}{v_0 + v_f}$     (6).

And finally with a little bit of arithmetic we could compute the final velocity as in:

$v_f = \sqrt{v_0^2 + 2 \cdot d \cdot g}$     (7).

We know all the quantities in the left side of (7), let’s plug the numbers in:

$v_f = \sqrt{12^2 \: \frac{m^2}{s^2} + 2 \cdot 40\: m \cdot 9.8 \: \frac{m}{s^2}}= \sqrt{928 \: \frac{m^2}{s^2}} = 30.46 \: \frac{m}{s} \approx 30.5 \: \frac{m}{s}.$

Knock out! After “only” 40 minutes (of yelling and cursing) we obtained the correct solution with a reasonable enough procedure.