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

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- schrodinger

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- anonymous

bad latex ._.

- Mimi_x3

lol fail

- Mimi_x3

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

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## More answers

- 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 \(\beta = tan^{-1} \frac{1}{3}\)

- Mimi_x3

better now :D

- Mimi_x3

|dw:1344003862684:dw|

- anonymous

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

- anonymous

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

- anonymous

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.

- anonymous

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

- anonymous

\[2\tan \beta = \tan \theta_1 + \frac{\sin \beta}{\cos \theta_1}\]
\[2\tan \beta = \tan \theta_2 + \frac{\sin \beta}{\cos \theta_2}\]
Solve it for \(\beta\).

- Mimi_x3

well thank you!

- anonymous

Did you get the answer?

- 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.

- anonymous

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')

- anonymous

angles*

- anonymous

i hope dumbcow comes up with much easier solution.

- Mimi_x3

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

- experimentX

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

- 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.

- anonymous

@Jemurray3 is there an easier way to do this?

- anonymous

Yes, give me just a second.

- 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.

- experimentX

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

- anonymous

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

- anonymous

why isn't link styling working?

- 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?

- anonymous

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?

- experimentX

well .. i guess \( 1.57 \approx 1.5 \)

- anonymous

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).

- anonymous

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.

- anonymous

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

- 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

- anonymous

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

- experimentX

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

- experimentX

* 1/3 = 0.3333

- experimentX

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

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