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  1. MHB Craftsman

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    #1
    A target is located at the distance 100 m and height 10 m. How much is the elevation angle of the shooter to be able to hit the target if the initial velocity is 60 m/s and g = $ \displaystyle 10m/s^2$?

    What I have done:
    100 = 60sin$ \displaystyle \alpha$t and 10 = 60sin$ \displaystyle \alpha$t - $ \displaystyle \frac{1}{2}(10)t^2$
    $ \displaystyle t=\frac{5cos\alpha}{3}$ and $ \displaystyle 10=t(60sin\alpha)-t)$
    Dunno what to do from here on.

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  3. Pessimist Singularitarian
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    #2
    Let's let $\theta$ be the angle of elevation. Let's orient our coordinate axes such that the vertical axis $y$ points up in the positive direction, and the horizontal axis $x$ points in the direction of motion in the positive direction.

    For the vertical component of motion, we have:

    $ \displaystyle a_y=-g$

    $ \displaystyle v_y=-gt+v_0\sin(\theta)$

    $ \displaystyle y=-\frac{g}{2}t^2+v_0\sin(\theta)t$

    For the horizontal component of motion, we have:

    $ \displaystyle a_x=0$

    $ \displaystyle v_x=v_0\cos(\theta)$

    $ \displaystyle x=v_0\cos(\theta)t$

    Let's eliminate the parameter $t$, by using $ \displaystyle t=\frac{x}{v_0\cos(\theta)}$:

    And so:

    $ \displaystyle y=-\frac{g}{2}\left(\frac{x}{v_0\cos(\theta)}\right)^2+v_0\sin(\theta)\left(\frac{x}{v_0\cos(\theta)}\right)=-\frac{g}{2v_0^2\cos^2(\theta)}x^2+\tan(\theta)x$

    Multiply through by $\cos^2(\theta)$:

    $ \displaystyle y\cos^2(\theta)=-\frac{g}{2v_0^2}x^2+\sin(\theta)\cos(\theta)x$

    Using double-angle identities, and multiplying through by 2, we may write:

    $ \displaystyle y\left(\cos(2\theta)+1\right)=-\frac{g}{v_0^2}x^2+\sin(2\theta)x$

    Arrange as:

    $ \displaystyle y+\frac{g}{v_0^2}x^2=\sin(2\theta)x-\cos(2\theta)y$

    Using a linear combination identity, we have:

    $ \displaystyle \sqrt{x^2+y^2}\sin\left(2\theta-\arctan\left(\frac{y}{x}\right)\right)=y+\frac{g}{v_0^2}x^2$

    Solving for $\theta$, there results:

    $ \displaystyle \theta=\frac{1}{2}\left(\arcsin\left(\frac{v_0^2y+gx^2}{v_0^2\sqrt{x^2+y^2}}\right)+\arctan\left(\frac{y}{x}\right)\right)$

    Now, all that's left is to plug in the given values.

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    #3
    Quote Originally Posted by Monoxdifly View Post
    A target is located at the distance 100 m and height 10 m. How much is the elevation angle of the shooter to be able to hit the target if the initial velocity is 60 m/s and g = $ \displaystyle 10m/s^2$?
    $\Delta x = v_0\cos{\theta_0} \cdot t \implies t = \dfrac{\Delta x}{v_0\cos{\theta_0}} = \dfrac{100}{60\cos{\theta}}$

    $\Delta y = v_0\sin{\theta_0} \cdot t - \dfrac{1}{2}gt^2 \implies 10 = 60\sin{\theta} \cdot \dfrac{5}{3\cos{\theta}} - 5\left(\dfrac{5}{3\cos{\theta}}\right)^2 \implies 10 = 100\tan{\theta} - \dfrac{125}{9} \sec^2{\theta}$

    using the identity $\sec^2{\theta} = 1 + \tan^2{\theta}$ yields a quadratic equation in $\tan{\theta}$ ...

    $0 = -43 + 180\tan{\theta} - 25\tan^2{\theta}$

    the quadratic formula yields two valid solutions for $\tan{\theta}$ ...

    $\tan{\theta} = \dfrac{18 \pm \sqrt{281}}{5}$

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    #4
    Yes, there should be two solutions, and in my post above, I should have included:

    $ \displaystyle \sqrt{x^2+y^2}\sin\left(\pi-\left(2\theta-\arctan\left(\frac{y}{x}\right)\right)\right)=y+\frac{g}{v_0^2}x^2$

    As a means of getting the other.

  6. MHB Craftsman

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    #5 Thread Author
    Quote Originally Posted by MarkFL View Post
    Let's let $\theta$ be the angle of elevation. Let's orient our coordinate axes such that the vertical axis $y$ points up in the positive direction, and the horizontal axis $x$ points in the direction of motion in the positive direction.

    For the vertical component of motion, we have:

    $ \displaystyle a_y=-g$

    $ \displaystyle v_y=-gt+v_0\sin(\theta)$

    $ \displaystyle y=-\frac{g}{2}t^2+v_0\sin(\theta)t$

    For the horizontal component of motion, we have:

    $ \displaystyle a_x=0$

    $ \displaystyle v_x=v_0\cos(\theta)$

    $ \displaystyle x=v_0\cos(\theta)t$

    Let's eliminate the parameter $t$, by using $ \displaystyle t=\frac{x}{v_0\cos(\theta)}$:

    And so:

    $ \displaystyle y=-\frac{g}{2}\left(\frac{x}{v_0\cos(\theta)}\right)^2+v_0\sin(\theta)\left(\frac{x}{v_0\cos(\theta)}\right)=-\frac{g}{2v_0^2\cos^2(\theta)}x^2+\tan(\theta)x$

    Multiply through by $\cos^2(\theta)$:

    $ \displaystyle y\cos^2(\theta)=-\frac{g}{2v_0^2}x^2+\sin(\theta)\cos(\theta)x$

    Using double-angle identities, and multiplying through by 2, we may write:

    $ \displaystyle y\left(\cos(2\theta)+1\right)=-\frac{g}{v_0^2}x^2+\sin(2\theta)x$

    Arrange as:

    $ \displaystyle y+\frac{g}{v_0^2}x^2=\sin(2\theta)x-\cos(2\theta)y$

    Using a linear combination identity, we have:

    $ \displaystyle \sqrt{x^2+y^2}\sin\left(2\theta-\arctan\left(\frac{y}{x}\right)\right)=y+\frac{g}{v_0^2}x^2$

    Solving for $\theta$, there results:

    $ \displaystyle \theta=\frac{1}{2}\left(\arcsin\left(\frac{v_0^2y+gx^2}{v_0^2\sqrt{x^2+y^2}}\right)+\arctan\left(\frac{y}{x}\right)\right)$

    Now, all that's left is to plug in the given values.
    Thank you! All comprehensible except the linear combination identity part. What's that and how was it derived?

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    #6
    Quote Originally Posted by Monoxdifly View Post
    Thank you! All comprehensible except the linear combination identity part. What's that and how was it derived?
    Here's a link that describes the derivation of the identity:


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