Math Forum game Yay!

[The Summoner of Leviathan]

if anyone cares for an easy-but-hard projectile problem i will be more than happy to bring it to the thread (even tho its a physics problem but lots of math are involved).

It has been a while but I can give it a shot.

[Doomie]

Mo never asked for a proof, simply a solution.

Here's a proof-type question for you.

Say I start walking at the Cartesian coordinate (0,0) and I need to go to (1,1). I can walk at right angles, going to (0,1) then (1,1) for a distance of 2, but it would be shorter to go at 45 degrees straight to (1,1), for a distance of 1.414. But say I move infinitesimally north, to (0,delta), then infinitesimally east, to (epsilon,delta), then repeat this in a staircase pattern. As epsilon and delta go to zero, this path should represent the 45 degree path from (0,0) to (1,1). However, as long as epsilon and delta are finite, I will have to walk a distance of 2. Is it a contradiction that my right-angle approximation of the 45 degree path still results in a distance of 2? Is the 45 degree path really shorter?

...yes.

[Mo-Nercy]

Mo never asked for a proof, simply a solution.

Yeah. Sorry guys. Bad wording on my part. Somewhere along the line I asked for a proof when all that was required was a solution.

*blush*

So embarassed. So much for being good at maths.

[eestlinc]

Mo never asked for a proof, simply a solution.

Here's a proof-type question for you.

Say I start walking at the Cartesian coordinate (0,0) and I need to go to (1,1). I can walk at right angles, going to (0,1) then (1,1) for a distance of 2, but it would be shorter to go at 45 degrees straight to (1,1), for a distance of 1.414. But say I move infinitesimally north, to (0,delta), then infinitesimally east, to (epsilon,delta), then repeat this in a staircase pattern. As epsilon and delta go to zero, this path should represent the 45 degree path from (0,0) to (1,1). However, as long as epsilon and delta are finite, I will have to walk a distance of 2. Is it a contradiction that my right-angle approximation of the 45 degree path still results in a distance of 2? Is the 45 degree path really shorter?

Yes the 45 degree path is shorter. As long as your minute stairsteps have a definite length, each individual staircase is an exact replica of the full coordinate square between (0,0) and (1,1) and each stairstep can be more quickly traversed by taking the hypotenuse rather than both sides of the stairstep. The smaller the size of each stairstep, the more stairsteps are necessary to cover the full distance, and although each tiny stairstep is closer in length to the actual hypotenuse, say perahps a difference of .00001, you will have to travel 100000 of these steps to arrive at the point (1,1).

Or do you want a more formal proof?

X = horizontal distance traveled = 1
Y = vertical distance traveled = 1
X+Y = 2
k = number of steps (let's say for convenience that each step is the same size, although it doesn't matter whether they are or not)

distance of each step horizontal = D_h
distance of each step vertical = D_v

D_h = X/k
D_v = Y/k

D_{total per step} = X/k + Y/k

D_interval = D_{total per step} * k

D_interval = k(X/k + Y/k) = k(X/k) + k(Y/k) = x+y = 2

this is true regardless of the value of K and is thus independent of the value of K (the number of steps)

therefore:

lim_k-oo D_interval = lim_k-oo (X+Y) = lim_k-oo (2) = 2.

[radyk05]

a projetile that is fired with an initial velocity v at an angle θ will pass through two points at hight h. show that, if the gun is set for maximun reach, the distance between the two points is

d = v/g * (v^2 - 4*g*h)^(1/2)

where g stands for the gravitational acceleration.
i'm going to make things a bit easier for those who haven't taken a physics or calculus course:
remember that we're dealing in two dimenssions so,
velocity in y, vy = sin θ * v
velocity in x, vx = cos θ * v

the equation for movement are:
x = xinitial + vx,initial * t
y = yinitial + vy,initial + (1/2) * g * t^2

where t stands for time.

that is all that you need.

have fun, i sure did.

[Del Murder]

Math looks so ugly typed out.

[radyk05]

Math looks so ugly typed out.

yes it does.

[Endless]

y = yinitial + vy,initial + (1/2) * g * t^2

I sense a typo.
Edit:
Try y = y₀ + v_y,0 * t - (1/2) * g * t²

Edit2:
While I'm at it:

v_x = dx/dt = v_x,0
v_y = dy/dt = v_y,0 - g * t

At the max height reached, v_y = 0, therefore t_{max height} = v_y,0/g
At that t_{max height}, half the horizontal distance is covered, so x_{max height} = x₀ + v_x,0 * v_y,0/g = sin θ * v * cos θ * v /g, which is max for θ = pi/4.
Now we can clean the mess, and rewrite:
x = x₀ + sqrt(2)/2 * v * t
y = y₀ + sqrt(2)/2 * v * t - (1/2) * g * t²

Now, let's take a point y_h.
y_h = y₀ + sqrt(2)/2 * v * t_h - (1/2) * g * t_h²
h = y_h - y₀
h = sqrt(2)/2 * v * t_h - (1/2) * g * t_h²

We solve this for t_h, we get:
t_h,2 = sqrt(2)/2g *(v + sqrt(v² - 4*h*g))
and
t_h,1 = sqrt(2)/2g *(v - sqrt(v² - 4*h*g))

d = x₂ - x₁ = sqrt(2)/2 * v * (t_h,2 - t_h,1)
d = sqrt(2)/2 * v * (sqrt(2)/2g *(v + sqrt(v² - 4*h*g)) - sqrt(2)/2g *(v - sqrt(v² - 4*h*g)))
d = v/g * sqrt(v² -4*h*g)

[theundeadhero]

Prove 2.9999999~ doesn't = 3

[Odaisé Gaelach]

Pi is exactly 3!

[Destai]

Math? *vomits*

("x=none of the above" is for wusses)

[radyk05]

*two golden stars for endless* (one is for fixing the equation)

[-N-]

Yes, eestlinc. Moreover, the "paradox" illustrates the fact that linear paths don't need to be approximated, and illustrates how problems arise as a result.

[eestlinc]

Prove 2.9999999~ doesn't = 3

or rather, prove that it does.

[Nick Schovitz]

Oh my God man my brain hurts! The college professor had to start this. I took that quiz and made a 42. Gosh nerds man!!!! I'm staying out of this thread now and beware that first question was just a warm up from Lyde and do you think that Quistis looks more like a librarian or a math teacher than a military teacher? Yes I do.