Thread: Newbie: Stuck on some Math

So, the formula is, rewritten as C-ish:

Code:

d = (v * cos(B) / g) * (sin(B) + sqrt(v * v * (sin(B) * sin(B) - 2 * g * h))

And you want to transform that into something that gives B = ....

That is beyond me - I'm just making sure that we're answering the right question first of all (one could easily spend quite a bit on rewriting the formula in different ways, and it's good to be sure that it's the RIGHT form first).

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Mats

And what does that formula represent (does it have a name?) - it would possibly be googleable if you have something to actually search for. But sin, cos, θ and square root are far too common on the Web to make it searchable (and of course it may not use θ).

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Mats

Originally Posted by filicudi

Well it's nothing common. It is the distance (d) traveled by a cannonball fired at velocity (v) at angle (θ) landing at a height (h).

Well, nothing common nowadays. I bet in medieval times that formula was all the rage.

QuantumPete

Ah, got it: It's a projectile trajectory calculation, right?

Note that the theta calculation would lead to a square equation - you get two different numerical answers (e.g. 30/60 degrees).

Edit: took too long to post!

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Mats

You need to improve your google skills:
http://en.wikipedia.org/wiki/Trajectory

Look for "angle of elevation", and it gives the formula you need.

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Mats

that would be the asin() function - there is also acos, atan.

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Mats

Originally Posted by matsp

Ah, got it: It's a projectile trajectory calculation, right?

Note that the theta calculation would lead to a square equation - you get two different numerical answers (e.g. 30/60 degrees).

Edit: took too long to post!

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Mats

Thats because one is considering a direct shot fired and the other angle is for a curved shot ( ie a mortar round's path of travel ). I would think for this problem you want the second, since a cannonball would travel as such, unless the distance was so minute that the shot didn't have time to begin its descent as gravity began to drag the ball down and the velocity of the shot slowed from friction, causing it to lose its ability to continue driving upward. Of course, your teacher may be treating this problem as a simple vacuum shot where neither gravity nor friction would come into play.

EDIT: whoops, sorry Mats...I didn't see your googled link which describes this much better than I could.

Does this help:

http://en.wikipedia.org/wiki/Traject...le#Height_at_x

Look for Angle theta required to hit coordinate ( x,y)