Solving the Complex Equation: d=2v2cos(o)sin(O+o)/gcos(O)

msadegian
Messages
8
Reaction score
0

Homework Statement


So you basically have to prove this big long equation d=2v2cos(o)sin(O+o)/gcos(O)


Homework Equations


sin(O+o)=sin(O)cos(o)+sin(o)cos(O)


The Attempt at a Solution



so i have weeded it down to this:

d=2v2((sinOsinO/cosOcoso)+(sinocosO/coso)+(sinOcoso/cosO)+(cosOcoso))/g

so i have the sin(O+o)=sin(O)cos(o)+sin(o)cos(O) part but i don't know how to get rid of the denominators and to get the other parts of the equation!
 
Physics news on Phys.org
msadegian said:

Homework Statement


So you basically have to prove this big long equation d=2v2cos(o)sin(O+o)/gcos(O)
Are you trying to prove that the equation is an identity, or are you trying to solve the equation? I don't think you're actually trying to prove this is an identity, but I have no idea what you're trying to solve for.

BTW, you're setting yourself up for disaster by choosing o and O for variables, both of which look a lot alike, and both of which resemble 0 (zero).

I would strongly advise changing your variables to something like this:
d=2v2cos(a)sin(b+a)/gcos(b)
msadegian said:

Homework Equations


sin(O+o)=sin(O)cos(o)+sin(o)cos(O)


The Attempt at a Solution



so i have weeded it down to this:

d=2v2((sinOsinO/cosOcoso)+(sinocosO/coso)+(sinOcoso/cosO)+(cosOcoso))/g

so i have the sin(O+o)=sin(O)cos(o)+sin(o)cos(O) part but i don't know how to get rid of the denominators and to get the other parts of the equation!
 
you're trying to prove the distance
and yeah well those were just the variables that were given to us but you're right it would be a lot easier to use a and b or something

any idea on how to solve?

basically the question he gave us is:
A cannon ball is fired from an angle o with an initial velocity of v. The hill slopes own with an angle of O. Prove that the horizontal distance the cannon ball travels is given by d=2v2cos(o)sin(O+o)/gcos(O)
 
msadegian said:
you're trying to prove the distance
No, you're not trying to prove the distance, which makes no sense. You're trying to prove that the distance can be obtained from the formula you showed. IOW, if the gun is fired as described, you have to come up with the formula d=2v2cos(a)sin(b+a)/gcos(b). You don't start from it; you end up with it.

How is this distance measured - horizontally or along the ground down the hill? It makes a difference.

You're going about this the wrong way. Based on the given information, you should end up with the formula for distance. What you seem to be doing is trying to use trig identities to write the formula in a different way. That is not what the problem is asking you to do.

Have you drawn a diagram? I can guarantee that you will have no success without a drawing.

What are the forces on the round after it leaves the barrel of the gun?
msadegian said:
and yeah well those were just the variables that were given to us but you're right it would be a lot easier to use a and b or something

any idea on how to solve?

basically the question he gave us is:
A cannon ball is fired from an angle o with an initial velocity of v. The hill slopes own with an angle of O. Prove that the horizontal distance the cannon ball travels is given by d=2v2cos(o)sin(O+o)/gcos(O)

Again, change the variables, with o = a, and O = b.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top