J-Rock physics from hell: The hide pitching a TV problem.

 

Legend has it that hide set a record for the longest distance achieved by a TV set pitched out a hotel room window. Now I’ve researched this and still have no idea if this record even exists, but I thought to myself…HEY!!! THIS WOULD MAKE A YAY BOUNCY PHYSICS PROBLEM!!!!! So true or not, prepare for suffering at the hands of meeeeeeeeeeee.

 

In this badly drawn windows-paint-made diagram, stick-figure-hide is launching Terebi-san (Trans. “Mr. TV”) set from a hotel room balcony (since I was too lazy to drawn a window.). I’ve assumed that his room is about 50 meters up from the ground. hide-sama pitches terebi-san at a 30 degree angle from the horizontal, and some seconds later, terebi-san smashes into the ground 180 m away and goes to TV heaven. Let’s find terebi-san’s initial velocity (the velocity with which hide-sama pitched him) and the time it took him to hit the ground. Neglect air resistance. I’m too lazy to solve it with air resistance. lol

The easiest way to solve this is to divide things up into two components: horizontal and vertical. First, take a close look at the velocity. Velocity is a vector, so it has direction. This means it’s going to have happy little components, one vertical component and one horizontal component:

Can you see how they sort of form a little triangle?

 

If you can remember all your trig, you’ll remember that   and  (where r is the hypotenuse)

From this you can derive the little values I have there for v_oy and v_ox

So anyway, now that you have the horizontal and vertical components of the velocity, make your self a little 2-column table with a “horizontal stuff” column and a “vertical stuff” column. List important variables such as time, displacement, velocity, and acceleration.

 

Notes about values:

* Notice I said “displacement” specifically instead of distance. We want the y amount of distance from where we started. The TV went up for a bit, then down and finally crashed on the ground 50 below the starting point (although not directly below), so the final displacement is 50 m below that point, or for calculations, -50.

** The time is the same for both horizontal AND vertical travel.

*** The acceleration acting on Terebi-san is acceleration due to gravity. Most people take this to be about 9.81. It’s negative because it’s pulling DOOOOOOWNward. (duh.)

 

Real quick, Here’s another bizarre little picture to help you visualize what exactly is going on with the velocity components and acceleration:

(Even though I can’t draw arrows to save my life, you should sort of be able to tell that the vertical velocity is changing because of the pull of gravity, while the horizontal velocity stays the same.

 

Anyway, now you’ll find your unknowns using kinematic equations.

 

The calculations for the horizontal are pretty easy.

Use the equation x = vt.

 x = vt

180 = .866v_ot

 

This may look hopeless, but it’s not. Setting up a certain equation for the vertical part will allow you to make substitutions back and forth.

For the vertical part, using the equation x = x_o + v_ot + ½ at² gives:

 

y = y_o + t – (9.81/2) t²

 

(remember that the “x” just refers to displacement. In this case, it’s displacement in the y (up and down) direction.)

Since you’re using the hotel room window as the point of origin, and dropping 50 m for your “y”, you can go ahead and take the initial displacement, y_o, to be 0. Now you have:

 

-50 = v_ot – (9.81/2) t²

 

It looks like we have two variables…But wait!!! We’ve got v_ot somewhere else too! Go back to the horizontal equation. Solve it for v_o and plug into the vertical equation:

 

180 = .866v_ot

v_o = 180/.866t = 207.85/t

                          

      -50 = v_ot – (9.81/2) t²

                     

                -50 = (207.85/t)t – (9.81/2) t²

And the t’s in the first term cancel so…

-50 = 207.85 – (9.81/2) t²

 

Now you can solve for t!

-257.85 = – (9.81/2) t²

t² = 52.569

t ≈ 7.25 s

 

That’s the time it takes terebi-san to smash into the ground. To get his initial velocity, go back and plug t into the x equation we had after solving for v_o.

 

v_o = 207.85/t

v_o = 207.85/7.25

v_o ≈ 28.7 m/s

 

Yay! Pointless answers to a pointless, relatively retarded problem. But hey maybe you learned something…-_- or not……..

Eh heh.

This is why I don’t teach physics…

 

OH OH OH OH !!!!!! But first…Here’s a fun continuation of the problem!!!!

 

Assume that hide-sama takes 1.5 s to actually accelerate the TV from rest before it leaves his hands to continue the journey down to crash land.

Given that the TV is 20 kg, figure out what force hide threw the TV with.

 

Here’s the magic equation, Newton’s second law: F = ma

We’re finding force, we know the mass…but what about acceleration? O_o

Well, acceleration is the change in velocity per unit time. (if you’ve had calculus: dv/dt)

The TV goes from 0 m/s to our previously found “initial” velocity of 28.7 m/s so that makes the change in velocity:

v - v_o = 28.7 – 0 = 28.7

Now it goes from 0 to 28.7 m/s in the space of 1.5 s, so that means that our change in time is 1.5.

So our acceleration is:

a = dv/dt = 28.7/1.5 ≈ 19.13 m/s²

 

NOW you can plug this into F = ma, along with the given mass of 20kg to get:

F = ma

F = 20 ( 19.13)

F ≈ 380 N

 

THAT’S ALL FOLKSSS!!!!!!

 

Back to pink spider!