Solving for Velocity Vector Angle of 45.0

In summary, the homework statement asks for when does the velocity vector make an angle of 45.0 with the x- and y-axes. The Attempt at a Solution found that the angle is when b/c is equal to the angle between the x- and y-axes.
  • #1
blufoggy
4
0

Homework Statement



If http://net.dinhweb.de/vector.gif ,[/URL] where b and c are positive constants, when does the velocity vector make an angle of 45.0 with the x- and y-axes?

Homework Equations



I'm not sure.. I've got a million equations but none of them seem to fit the problem the way I need. I know this isn't relevant to the question, but the prompt asks for "t=?"

The Attempt at a Solution



I've got this thing worked down to b divided by c (b/c), but apparently my "answer is off by a multiplicative factor." I've already taken a few shots at guessing the coefficient I need, but no luck there, so I decided to give this a try.

I didn't think any kind of "multiplicative factor" was necessary because equal "i hat" and "j hat" values would give a 45 degree angle. Apparently not and here I am stuck on this problem

Any kind of help would be greatly appreciated =]
 
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  • #2
Do you know how to use the scalar product to determine the angle between two vectors? i.e.[itex]\bold{x}\cdot \bold{y}=xy\cos\theta[/itex] where [itex]\theta[/itex] is the angle between the two vectors.
 
  • #3
Somewhat.. Researching the answer earlier through my textbook actually led me to the exact page with the scalar product definition, but I couldn't figure out how it tied into the problem.

It's actually a couple chapters back from the original problem, but now I'm in the process of reading it over again.
 
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  • #4
Well, try looking at r.i and r.j, since you know the angle that r must make with the x and y axes.
 
  • #5
Hm.. I think I understand a little better now..

So I have the component direction and magnitude, but finding the dot product of r and i or r and j gives me the magnitude of r which is the component of either i or j? And that magnitude is the factor I'm missing?
 
  • #6
I have looked at this, and plugged b/c into excel and it works for 5 different b and c constants. Thus for me, b/c is correct.
 
  • #7
Well I feel stupid..

Someone hinted to "find the derivative of the positive vector".. Which meant next to nothing to me, but I decided to give it a try anyway and use the coefficients I would've gotten if I took the derivative of the whole equation. It turns out the "mutiplicative factors" I was missing was a 2 and 3, so the answer was..

t = (2b)/(3c)

I don't know if it's just me, but that answer doesn't make any sense and now I'm more confused than ever.
 

1. What is the formula for solving for velocity vector angle of 45.0?

The formula for solving for velocity vector angle of 45.0 is given by tan(θ) = vy/vx, where θ is the angle of the velocity vector and vy and vx are the vertical and horizontal components of the velocity vector, respectively.

2. How is the velocity vector angle of 45.0 related to projectile motion?

The velocity vector angle of 45.0 is the angle at which a projectile is launched in order to achieve maximum range, assuming no air resistance. This angle is known as the optimal launch angle for projectile motion.

3. Why is the velocity vector angle of 45.0 significant?

The velocity vector angle of 45.0 is significant because it represents the maximum possible range for a projectile launched at a given initial velocity. This angle also allows for symmetric motion, meaning the projectile reaches the same height at its peak as it does at its launch point.

4. How can the velocity vector angle of 45.0 be used to solve for other projectile motion variables?

The velocity vector angle of 45.0 can be used to solve for other projectile motion variables such as the initial velocity, maximum height, and time of flight. By knowing the angle at which the projectile is launched, the equations for projectile motion can be used to determine these other variables.

5. Can the velocity vector angle of 45.0 be used for all types of projectile motion?

No, the velocity vector angle of 45.0 is only applicable for projectile motion in a vacuum, where there is no air resistance. In real-world situations, air resistance will affect the trajectory of the projectile, and the optimal launch angle may be different depending on the specific conditions.

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