A physics proof!

[struntz]

Hello everyone!

I was wondering if anyone could prove this:
Range = Vo^2 sin 2x/g
Note: g is acceleration due to gravity
Its highschool physics by the way!
Thanks for listening

[krtxmrtz]

Originally posted by struntz
Hello everyone!

I was wondering if anyone could prove this:
Range = Vo^2 sin 2x/g
Note: g is acceleration due to gravity
Its highschool physics by the way!
Thanks for listening

What is x? What is v₀? Range of what? Is g dividing x or the entire right hand side? In either case, you must have a non-dimensional quantity as the angle you calculate the sinus of.

[bugzpodder]

v0 is probably v₀ which is used to mean initial velocity. anywayz what is this "range" used for? range of what???

[krtxmrtz]

Originally posted by bugzpodder
v0 is probably v₀ which is used to mean initial velocity. anywayz what is this "range" used for? range of what???

I'd guessed v₀ might be initial velocity but a range is usually a distance so there seems to be yet another inconsistency here...

[krtxmrtz]

You know what? I'd bet the whole confusion comes from a transcription error. If the formula is written as:

Range = v₀ Sqrt (2x/g)

then it does make sense, as least as far as units go.

For example,

(cm/s) Sqrt(cm/(cm/s²)) = (cm/s) sqrt(s²) = (cm/s) s = cm

If that's the case, then the formula is very easy to prove based on basic physics laws.

[A$$Bandit]

For those who don't know, he/she's asking for the range of a projectile thrown at 'u ms^-1', at an angle of x^o to the horizontal. 'g' is the acceleration downwards due to gravity, and is usually taken to equal 9.8 ms^-2.

Using the formula "s = ut + at²/2" in the vertical direction (up not down), to find out when it hits the ground again, you get:

0 = u.sinx.t - gt²/2
t(gt/2 - u.sinx) = 0
t = 0 (when u threw it)
or
t = 2u.sinx / g (when it hits the ground again)

Now you can plug that into "s = ut + at²/2" but in the horizontal direction. Since a = 0 in that direction the formula can be reduced to "s = ut":

Range = u.cosx((2usinx)/g)
Simplify it:
Range = 2u²sinx.cosx / g

Finally, using the formula "sin2x = 2sinx.cosx", that simplifies to:

Range = 2u²sin2x / g

If that still doesn't help, go HERE

[mlewis]

uh, no. g = 9.81 m/s2 not m/s^-2 LOL... m/s^-2 would be ms^2

[A$$Bandit]

Don't you wink at me! I am right to say that an approximation for 'g' is 9.8 ms^-2, and if you don't understand why then I suggest you get some help!

[mlewis]

Doh, never mind

[struntz]

Hey A$$Bandit, thanks for the responce, I just have one question....is u, Initial Velocity? like when you wrote
u.sinx.t - gt2/2, is that
Vo(sinx)(t)-g(t)^2
-----------------------
2
or is it
Vo(sinx)(t)-g(2t)
---------------------
2

[bugzpodder]

although this question is not directed at me, but i can tell you that u is the initial velocity and g is the acceleration due to gravity=9.8 m/s²

the equation of height is:

h=u*sinx*t - g*(t²)/2

where x is the angle and t is the time in seconds.

[A$$Bandit]

Yeh it is. Dunno about everywhere else, but in England we use 'u' for initial velocity.

[opus]

QUOTE]Dunno about everywhere else, but in England we use 'u' for initial velocity.[/QUOTE]
Yes, but you're on an island, still using a currency that goes by Kg or pounds or what?