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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 \(\beta = tan^{1} \frac{1}{3}\)
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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Mimi_x3
 3 years ago
Best ResponseYou've already chosen the best response.1something is wrong with this site; its not suppose to fail

Mimi_x3
 3 years ago
Best ResponseYou've already chosen the best response.1A 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 \(\beta = tan^{1} \frac{1}{3}\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[x = u \cos \theta \cdot t\] \[y = u\sin \theta \cdot t  \frac12\cdot g\cdot t^2\] \[v_y u_y = gt \implies u\sin \beta  u\sin \theta =gt \implies \frac{u\sin \beta u\sin \theta}{g}\] \[\frac{y}{x} = \tan \beta\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[t = \frac{u\sin\theta  \sin \beta}g\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0From 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[2\tan \beta={\tan \theta} + \frac{\sin \beta}{\cos\theta}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[2\tan \beta = \tan \theta_1 + \frac{\sin \beta}{\cos \theta_1}\] \[2\tan \beta = \tan \theta_2 + \frac{\sin \beta}{\cos \theta_2}\] Solve it for \(\beta\).

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Did you get the answer?

Mimi_x3
 3 years ago
Best ResponseYou've already chosen the best response.1haven'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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hmm 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')

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i hope dumbcow comes up with much easier solution.

Mimi_x3
 3 years ago
Best ResponseYou've already chosen the best response.1well the solution that i have uses the sum of two roots.. oh forget it

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1case 1) dw:1344008942118:dw case 2)dw:1344009010791:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1assume 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Jemurray3 is there an easier way to do this?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, give me just a second.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1the first will give you the value of time ... use that time in to find out the coordinates ... then find the elevation as in second.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.11) http://www.wolframalpha.com/input/?i=solve+%28x+cos+45%29+%3D+%28y+cos+63.26%29%2C+%28a+sin+63.26+x++1%2F2+9.8*+x^2%29+%3D+%28a+sin+45+y++1%2F2+9.8*+y^2%29 2) 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\href{ http://www.wolframalpha.com/input/?i=%280.0476134%29%2F%280.0302971%29 }{\text{y/x should be tan, right?}}\]dw:1344012709905:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0why isn't link styling working?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1yep ... 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0With g=10. http://www.wolframalpha.com/input/?i=%28.0466611%29%2F%28.0296912%29 http://www.wolframalpha.com/input/?i=solve+%28x+cos+45%29+%3D+%28y+cos+63.26%29%2C+%28a+sin+63.26+x++1%2F2+10*+x%5E2%29+%3D+%28a+sin+45+y++1%2F2+10*+y%5E2%29 It still isn't 1/3. Is the question wrong?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1well .. i guess \( 1.57 \approx 1.5 \)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0This 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).

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Lastly, 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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1lol 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh, so you did get the answer @experimentX eyeopening solution @Jemurray3

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1sorry ... i forgot the answer was 1/3

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1sorry for earlier ... i had been putting angles oppositely http://www.wolframalpha.com/input/?i=solve+%28x+cos+45%29+%3D+%28y+cos+63.26%29%2C+%28a+sin+45+x++1%2F2+9.8*+x%5E2%29+%3D+%28a+sin+63.26+y++1%2F2+9.8*+y%5E2%29 the answer would be 1/3 http://www.wolframalpha.com/input/?i=arctan%28%28sin+45+0.0966941++1%2F2+*+9.8*0.0966941%5E2%29%2F%28cos%2845%29*0.0966941%29%29 still not quite accurate
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