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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 \(\beta = tan^{1} \frac{1}{3}\)
 one year ago
 one year 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 \(\beta = tan^{1} \frac{1}{3}\)
 one year ago
 one year ago

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Mimi_x3Best ResponseYou've already chosen the best response.0
something is wrong with this site; its not suppose to fail
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.0
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 \(\beta = tan^{1} \frac{1}{3}\)
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.1
\[x = u \cos \theta \cdot t\] \[y = u\sin \theta \cdot t  \frac12\cdot g\cdot t^2\] \[v_y u_y = gt \implies u\sin \beta  u\sin \theta =gt \implies \frac{u\sin \beta u\sin \theta}{g}\] \[\frac{y}{x} = \tan \beta\]
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.1
\[t = \frac{u\sin\theta  \sin \beta}g\]
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.1
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.
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.1
\[2\tan \beta={\tan \theta} + \frac{\sin \beta}{\cos\theta}\]
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.1
\[2\tan \beta = \tan \theta_1 + \frac{\sin \beta}{\cos \theta_1}\] \[2\tan \beta = \tan \theta_2 + \frac{\sin \beta}{\cos \theta_2}\] Solve it for \(\beta\).
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.1
Did you get the answer?
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.0
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.
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.1
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')
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.1
i hope dumbcow comes up with much easier solution.
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.0
well the solution that i have uses the sum of two roots.. oh forget it
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
case 1) dw:1344008942118:dw case 2)dw:1344009010791:dw
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
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.
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.1
@Jemurray3 is there an easier way to do this?
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.2
Yes, give me just a second.
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
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.
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
1)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
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.1
\[\href{http://www.wolframalpha.com/input/?i=%280.0476134%29%2F%280.0302971%29}{\text{y/x should be tan, right?}}\]dw:1344012709905:dw
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.1
why isn't link styling working?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
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?
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.1
With 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?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
well .. i guess \( 1.57 \approx 1.5 \)
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.2
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).
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.2
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.
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.2
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
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
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
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.1
Oh, so you did get the answer @experimentX eyeopening solution @Jemurray3
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
sorry ... i forgot the answer was 1/3
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
sorry 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
 one year ago
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