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Mimi_x3

  • 3 years ago

A cannon shell is projected at \(u\) m/s at elevation angle at \(\theta\). The cannon hit the target at \(X,Y)\) when fired at two different angles \(45\) degrees and \(63°26'\). Show that the angle of elevation of the target from the gun is given by \(\beta = tan^{-1} \frac{1}{3}\)

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  1. Ishaan94
    • 3 years ago
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    bad latex ._.

  2. Mimi_x3
    • 3 years ago
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    lol fail

  3. Mimi_x3
    • 3 years ago
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    something is wrong with this site; its not suppose to fail

  4. Mimi_x3
    • 3 years ago
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    A cannon shell is projectile at \(u\) m/s at elevation angle at \(\theta\). The cannon hit the target at (\(X,Y)\) when fired at two different angles \(45\) degrees and \(63°26'\). Show that the angle of elevation of the target from the gun is given by \(\beta = tan^{-1} \frac{1}{3}\)

  5. Mimi_x3
    • 3 years ago
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    better now :D

  6. Mimi_x3
    • 3 years ago
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    |dw:1344003862684:dw|

  7. Ishaan94
    • 3 years ago
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    \[x = u \cos \theta \cdot t\] \[y = u\sin \theta \cdot t - \frac12\cdot g\cdot t^2\] \[v_y -u_y = -gt \implies u\sin \beta - u\sin \theta =-gt \implies \frac{u\sin \beta -u\sin \theta}{g}\] \[\frac{y}{x} = \tan \beta\]

  8. Ishaan94
    • 3 years ago
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    \[t = \frac{u\sin\theta - \sin \beta}g\]

  9. Ishaan94
    • 3 years ago
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    From the first two equations and the last one. \[\frac{y}x = \frac{u\sin \theta\cdot t-\frac12gt^2}{u \cos \theta\cdot t} = \tan \theta - \frac12\cdot g\cdot\frac{t}{u\cos \theta}\] Substitute the expression for t.

  10. Ishaan94
    • 3 years ago
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    \[2\tan \beta={\tan \theta} + \frac{\sin \beta}{\cos\theta}\]

  11. Ishaan94
    • 3 years ago
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    \[2\tan \beta = \tan \theta_1 + \frac{\sin \beta}{\cos \theta_1}\] \[2\tan \beta = \tan \theta_2 + \frac{\sin \beta}{\cos \theta_2}\] Solve it for \(\beta\).

  12. Mimi_x3
    • 3 years ago
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    well thank you!

  13. Ishaan94
    • 3 years ago
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    Did you get the answer?

  14. Mimi_x3
    • 3 years ago
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    haven't tried it; looks complicated than i thought, might try it later. i already have the solution but the method looked complicated; so i tried posting it here if there is an easier method; i guess not..oh wells.

  15. Ishaan94
    • 3 years ago
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    hmm i don't think my equations are wrong but i am still suspecting something's not right with my solution. i wish angle's were easy to calculate (arctan1/3 and 63*26')

  16. Ishaan94
    • 3 years ago
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    angles*

  17. Ishaan94
    • 3 years ago
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    i hope dumbcow comes up with much easier solution.

  18. Mimi_x3
    • 3 years ago
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    well the solution that i have uses the sum of two roots.. oh forget it

  19. experimentX
    • 3 years ago
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    case 1) |dw:1344008942118:dw| case 2)|dw:1344009010791:dw|

  20. experimentX
    • 3 years ago
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    assume that case 1) arrives at the point in time 't1' and case 2) arrives at time 't2' equate those two cases, |dw:1344009086029:dw| Two unknowns ... two equations ... solve for t1 and t2 ... (find at least one value). and put those in your equation to get the coordinate of x and y ... now you have coordinates ... find the angle of elevation.

  21. Ishaan94
    • 3 years ago
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    @Jemurray3 is there an easier way to do this?

  22. Jemurray3
    • 3 years ago
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    Yes, give me just a second.

  23. experimentX
    • 3 years ago
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    the first will give you the value of time ... use that time in to find out the coordinates ... then find the elevation as in second.

  24. Ishaan94
    • 3 years ago
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    \[\href{ http://www.wolframalpha.com/input/?i=%280.0476134%29%2F%280.0302971%29 }{\text{y/x should be tan, right?}}\]|dw:1344012709905:dw|

  25. Ishaan94
    • 3 years ago
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    why isn't link styling working?

  26. experimentX
    • 3 years ago
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    yep ... it's close. moreover i used lots of approximations ... and got pretty close. i guess there should be better solution that would give accurate solution ... BTW what is the value of g?

  27. experimentX
    • 3 years ago
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    well .. i guess \( 1.57 \approx 1.5 \)

  28. Jemurray3
    • 3 years ago
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    This needs to be true for all velocities and accelerations so for convenience let's go ahead and set the velocity to 1 and the acceleration to -2. That yields the following: \[x = t\cos(\theta)\] \[y = t\sin(\theta)-t^2\] Solving the first equation for t and plugging it in to the second, \[ y = x\tan(\theta) - x^2\sec^2(\theta) \] We are asked to show that \[y = \frac{1}{3} x\] so we need to show that \[x\left( \tan(\theta) - \frac{1}{3} \right) - x^2\sec^2(\theta) = 0\] for the same point x and the two different angles above. This problem is just the intersection of two parabolas. Dividing out an x and solving yields \[x = \frac{\tan(\theta) - \frac{1}{3}}{\sec^2(\theta)} \] just plug in and verify that this quantity is the same for both angles (it is).

  29. Jemurray3
    • 3 years ago
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    You could do it with v and g left as unspecified constants but in the end you would just get the above expression multiplied by 2v^2/g ,which would still remain unchanged.

  30. Jemurray3
    • 3 years ago
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    Lastly, if you would prefer to pretend that you didn't know the answer ahead of time, you could substitute some variable alpha rather than 1/3, and solve like this: http://www.wolframalpha.com/input/?i=%28+tan%28x%29-+alpha%29+%2F+sec^2%28x%29+%3D+%28+tan%28y%29+-+alpha%29+%2F+sec^2%28y%29+where+x%3D+45+degrees+and+y+%3D+63+degrees+26+arc+minutes

  31. experimentX
    • 3 years ago
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    lol what have i been doing ... @Ishaan94 that you are taking is ratio of time ... you need to plugin into x(t) and y(t) to get the value http://www.wolframalpha.com/input/?i=arctan%28%28sin+45+0.0302971+-+1%2F2+*+9.8*0.0302971^2%29%2F%28cos%2845%29*0.0302971%29%29

  32. Ishaan94
    • 3 years ago
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    Oh, so you did get the answer @experimentX eye-opening solution @Jemurray3

  33. experimentX
    • 3 years ago
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    sorry ... i forgot the answer was 1/3

  34. experimentX
    • 3 years ago
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    * 1/3 = 0.3333

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