The Random Thread

[anoni]

It's just big numbers, the only equation that was used was the conservation of momentum

(m1 * v1 = m2 * v2)

So it's not hard, just big numbers (and I used a calculator for the big numbers xD)

[White Wolf Guardian]

Srs posts, nope. Why not a lovely dose of /k/?

[drenteɳ]

Please anoni, no more physics.......... I've had enough for this year!!!!!! (exam's over, physics is done)
I wanted a break!

[anoni]

NOW CLASS

I fired a 4 kg projectile out of a cannon that had a force of 60 N, I fired it at an angle 50 degrees to the horizontal on a flat surface. Assuming no air resistance or other effects, and take acceleration due to gravity to be 9.8 m/s² how far would the projectile go?

MWHAHAHAHAHAHAHAA *evil laugh*

[White Wolf Guardian]

Nope. * Sends Anoni to challenge the telekinetic Episode of Stan Lee's show*

[George]

One who cuddles much will appear very soon...

[White Wolf Guardian]

*Cuddles the Billy dragon*

[drenteɳ]

Hang on anoni...... don't I need to know how long the force is applied for, or did I miss something???

[anoni]

Assume the force was applied for one second

[Luggz]

When you assume, you make an ass out of u and me

[drenteɳ]

Wow.... that must be a long barrel on the cannon.


4kg 60N 50degrees 1s


60 = 4a      a = 15ms^-2     15 * 1 = 1     v = 15ms^-1              That means that the barrel is 7.5 meters long, assuming force is constant along it and has an instant dropoff at the end. 

Anyway, now in the vertical direction, the velocity is sin(50) * 15 and in the horizontal, it's cos(50) * 15, because triangles are triangles. What we care about for now is the vertical velocity (11.491m/s). I'm assuming that the bottom of the barrel is touching the ground, so the top is at sin(50) * 7.5 meters (5.745m).


With these, I know that the peak of the flight will be reached at 11.491/9.8 seconds (1.173s) and will be 5.745*1.173 meters above the end of the barrel (12.484m above the ground).
Because v² = u² + 2ax, the velocity at the ground level is equal to the square root of 2ax (u = 0) and 2ax = 244.686, so v = 15.642m/s.


Time to ground = time to accelerate to 15.642m/s = 15.642 / 9.8 = 1.596s
The total time in the air is 1.173 + 1.596 seconds (2.769 seconds).




Now, to figure out the distance travelled, we can simply say that the horizontal velocity (9.642m/s) times the time (2.769s) is the distance. This gives us a distance of 26.699m.
Am I right?????? I hope I didn't screw up (I know I could have saved some equations if I looked up formulas, but you know... fun!)

[The Past]

I don't even! I DON'T! And now I must sleep.

[anoni]

Yep, looks like you got everything right.

I mean, originally I was assuming that the barrel was lengthless but the added complexity kind of made the question funner so yeah

Using the added complexity, my method was to simply solve the quadratic

0 = -a*t^2 + bt + c

where a = (1/2)g,
b = v sin(50) [v = 15]
c = l sin(50) [l = 7.5]

this would give you the full time but there is a whole range  of different methods one could try.

Good job! You may be the first person to actually correctly do one of my maths/physics questions AND you made it more complicated than I anticipated, so pat on the back for you

Wow.... that must be a long barrel on the cannon.


4kg 60N 50degrees 1s


60 = 4a      a = 15ms^-2     15 * 1 = 1     v = 15ms^-1              That means that the barrel is 7.5 meters long, assuming force is constant along it and has an instant dropoff at the end. 

Anyway, now in the vertical direction, the velocity is sin(50) * 15 and in the horizontal, it's cos(50) * 15, because triangles are triangles. What we care about for now is the vertical velocity (11.491m/s). I'm assuming that the bottom of the barrel is touching the ground, so the top is at sin(50) * 7.5 meters (5.745m).


With these, I know that the peak of the flight will be reached at 11.491/9.8 seconds (1.173s) and will be 5.745*1.173 meters above the end of the barrel (12.484m above the ground).
Because v² = u² + 2ax, the velocity at the ground level is equal to the square root of 2ax (u = 0) and 2ax = 244.686, so v = 15.642m/s.


Time to ground = time to accelerate to 15.642m/s = 15.642 / 9.8 = 1.596s
The total time in the air is 1.173 + 1.596 seconds (2.769 seconds).




Now, to figure out the distance travelled, we can simply say that the horizontal velocity (9.642m/s) times the time (2.769s) is the distance. This gives us a distance of 26.699m.
Am I right?????? I hope I didn't screw up (I know I could have saved some equations if I looked up formulas, but you know... fun!)

[drenteɳ]

Oh..... I completely missed the fact that I could do that lol.
Just a distance/time graph equation.


I think my time calculation is probably a few milliseconds off due to rounding, but yeah close enough.






My turn to give questions. A motorbike is at rest 5m away from a 10m long ramp at a 30 degree angle. An identical ramp is set up facing the opposite direction with a 20m gap between the two ramps. If the motorbike  provides  4900N of force while in contact with the ground and has a mass of 700kg, will it make the jump? Ignore friction, air resistance, whatever else. Assume this is on Earth and not a planet with different gravity.

[Tim Siguire]

Wow 2 people know physics. I thought of something physics like but It's more like a theoretical question.




*Stabs Sword Deep Into the earth*Only they whom is deemed worthy can wield this fine Holoblade.




(Im sorry if i messed anything or all of it. Physics sounds cool and I would like to learn it but I have difficulties understanding complex equations)




Ok, say you had an aerial vehicle, say a commercial jet
and you had a double-barred cannon built into the aircraft. (let's go with the Hull)
If the Jet was over 2.5 tons (5,000 Lbs) and was falling at 1400 m/s at a 65º angle facing earth
plus the applied force of gravity 9.8 m/s^2.
Including we know the cannon's both barrels fire at the same time.
The cannon shoots at a force of 240 N.
Then how many shots are needed to slow the jet to a safe landing?
The jet low enough to crashland and be destroyed, about 130,000 M up.




5,000 Lbs - 240 N - 9.8 m/s^2 - 1400 m/s


Do mine whenever you want lol.
Is this possible? Is everything there? am I missing anything?