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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

bad latex ._.

2. Mimi_x3

lol fail

3. Mimi_x3

something is wrong with this site; its not suppose to fail

4. Mimi_x3

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

better now :D

6. Mimi_x3

|dw:1344003862684:dw|

7. Ishaan94

$x = u \cos \theta \cdot t$ $y = u\sin \theta \cdot t - \frac12\cdot g\cdot t^2$ $v_y -u_y = -gt \implies u\sin \beta - u\sin \theta =-gt \implies \frac{u\sin \beta -u\sin \theta}{g}$ $\frac{y}{x} = \tan \beta$

8. Ishaan94

$t = \frac{u\sin\theta - \sin \beta}g$

9. Ishaan94

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

$2\tan \beta={\tan \theta} + \frac{\sin \beta}{\cos\theta}$

11. Ishaan94

$2\tan \beta = \tan \theta_1 + \frac{\sin \beta}{\cos \theta_1}$ $2\tan \beta = \tan \theta_2 + \frac{\sin \beta}{\cos \theta_2}$ Solve it for $$\beta$$.

12. Mimi_x3

well thank you!

13. Ishaan94

Did you get the answer?

14. Mimi_x3

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

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

angles*

17. Ishaan94

i hope dumbcow comes up with much easier solution.

18. Mimi_x3

well the solution that i have uses the sum of two roots.. oh forget it

19. experimentX

case 1) |dw:1344008942118:dw| case 2)|dw:1344009010791:dw|

20. experimentX

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

@Jemurray3 is there an easier way to do this?

22. Jemurray3

Yes, give me just a second.

23. experimentX

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. experimentX
25. Ishaan94

$\href{ http://www.wolframalpha.com/input/?i=%280.0476134%29%2F%280.0302971%29 }{\text{y/x should be tan, right?}}$|dw:1344012709905:dw|

26. Ishaan94

why isn't link styling working?

27. experimentX

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?

28. Ishaan94
29. experimentX

well .. i guess $$1.57 \approx 1.5$$

30. Jemurray3

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).

31. Jemurray3

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.

32. Jemurray3

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

33. experimentX

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

34. Ishaan94

Oh, so you did get the answer @experimentX eye-opening solution @Jemurray3

35. experimentX

sorry ... i forgot the answer was 1/3

36. experimentX

* 1/3 = 0.3333

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