Recent content by squid

All clear so far. What next? [EDIT] Never mind. I did some reading and solved it on my own. The first derivative of the partial expression above is '-cos(2*a - b)*cos(b)*csc(a-b)^2*sec(a)^2', so to find the critical points I just had to solve: -cos(2*a - b)*cos(b)*csc(a-b)^2*sec(a)^2 = 0...

Could you please post your calculations? I'll try to figure them out with this calculus book I found on my dad's bookshelf... (Yes, I WILL figure it out... After all, they don't teach us trigonometry in 9th grade either and I did figure that out...)

Yep. It corresponds to the results from the graphs. Thanks! But how did you come up with it?

I'll check if it works. Thanks. :)

Is 'a' the angle of minimal speed?

Nope, but let's assume that I do. How does that help me? [EDIT] Isn't there a simple way to solve this problem?? [EDIT 2] This is actually a parabola... There must be a formula to find the minimum of a parabola, right...?

Nope. Anyway, according to multiple graphs, the optimal angle is somewhere near 0.8 radians, that's somewhere near 45 degrees... The graphs are, of course, inaccurate, but I bet it is exactly 45 degrees. I thought so from the very beginning, but it seems kind of odd... Some points can't be...

How should I know? I'm a 9th grade student... (Who really sucks at math!) [EDIT] The only way I can think of is using a graph, but that doesn't help me to write a formula...

Hi! I was given this physics problem and I'm trying to solve it, but I'm not sure if I'm doing it correctly... I have to write a formula to calculate the _minimal_ speed (I missed this part at first) at which a projectile must be shot to pass through point (x, y). My first step was to write...

Hi! Could anyone explain me why Velocity Verlet integration works and how did Loup Verlet come up with it? Thanks!

Hi! I have a simple question: How does the friction force that acts on a body that is laying on an inclined surface depend on the angle of the surface? Thanks.

So the solution to my equation is: s = +- (sqrt(g) * (x - x0) * sec(a)) / (sqrt(2) * sqrt(tan(a) * (y - y0)^2 - y + y0)) Is that correct? [EDIT] I get 42.3 as well! I guess it is correct, then! Your idea of making 'g' positive was brilliant! Thanks! :D

That's what I was talking about when I said "I tried solving it the other way around". Test it for origin point '(0, 0)', destination point '(100, 10)', angle '0.4' and gravitational acceleration of '-9.8'. You get '-32.2793'. [EDIT] If I use a negative angle, the solution is positive...

Given a projectile at position '(x0, y0)' and a launch angle 'a', at which speed must I launch the projectile to make it pass through a point '(x, y)', assuming that the only force that acts upon it is gravity (which causes acceleration 'g')? I came up with the following: vx0 = s * cos(a)...