Optimizing Distance from a Cannon: Simplifying the Quadratic Formula

  • Thread starter Thread starter kalpalned
  • Start date Start date
  • Tags Tags
    Cannon
AI Thread Summary
The discussion revolves around optimizing the angle for maximum distance from a cannon using the quadratic formula. The correct angle is identified as theta = 70.5 degrees, but the user questions whether there's a simpler method to avoid complex algebra. It is clarified that solving the quadratic is unnecessary since the range (r) is always increasing, indicating no solutions exist for certain parameter conditions. The conversation emphasizes understanding the relationship between parameters rather than focusing solely on algebraic solutions. Overall, the thread highlights the importance of recognizing when algebraic solutions may not be applicable in optimization problems.
kalpalned
Messages
2
Reaction score
0

Homework Statement


fullsizeoutput_92.jpeg

Homework Equations


Hint given in problem statement
Quadratic formula

The Attempt at a Solution


fullsizeoutput_93.jpeg

The correct answer is theta = 70.5 degrees. I'd like to know if I'm on the right track. Do I simply solve for theta? Is there a better approach that circumvents all of this messy algebra? Much appreciated!
 
Physics news on Phys.org
Oops! All wrong. My bad. Please ignore
 
Last edited:
kalpalned said:
Is there a better approach that circumvents all of this messy algebra?
You do not need to solve the quadratic. If r is always increasing the quadratic has no solutions. What condition on the parameters leads to that?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top