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Thread: max range of projectile

  1. #1
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    max range of projectile

    a particle is projected up an inclined plane with initial velocity u m/s .the line of projection makes angle a with the horizontal and the inclined plane makes angle b with the horizontal. show that the range is a maximum when a=45+b/2

    so far I got the time of flight =  2usinb-a/(gcosa)
    and i have gotten the range =  [cos(b-a)   -{(sina*sinb-a)/cosa} ] [(2u^2sinb-a)/gcosa]
    from here i cant see how to maximise this using differentiation or trigonometry.
    Last edited by markosheehan; Jun 27th 2017 at 01:52 AM.
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  2. #2
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    Re: max range of projectile

    if you can not understand my brackets, sorry and please ask me to explain what you do not get
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  3. #3
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    Re: max range of projectile

    Quote Originally Posted by markosheehan View Post
    a particle is projected up an inclined plane with initial velocity u m/s .the line of projection makes angle a with the horizontal and the inclined plane makes angle b with the horizontal. show that the range is a maximum when a=45+b/2

    so far I got the time of flight =  2usinb-a/(gcosa)
    and i have gotten the range =  [cos(b-a)   -{(sina*sinb-a)/cosa} ] [(2u^2sinb-a)/gcosa]
    from here i cant see how to maximise this using differentiation or trigonometry.
    Since the problem is to show that "a= 45+ b/2", a is the variable and you can set the derivative, with respect to a, equal to 0.
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  4. #4
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    Re: max range of projectile

    there is more than one variable (a and b) so i do not see how i can do this.
    I am trying to get a as one term and then differentiate it.
    i have gotten here : [cosa * cosb-a -(sinasinb-a)] [(2u^2 sinb-a)/g]
    and then using the trig formula cos(A+B) = cosAcosB− sinAsinB
    cos a+b-a [(2u^2 sinb-a)/g]
    cos b [(2u^2 sinb-a)/g]
    I can not get the answer
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  5. #5
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    Re: max range of projectile

    show that the range is a maximum when a=45+b/2
    note incline angle $b$ may be treated as a constant ... angle $a$ is variable.

    Let the surface of the incline be the x-axis, y-axis perpendicular to the x-axis, $(x_0,y_0) = (0,0)$ ... see attached diagram

    $x = u\cos(a-b) \cdot t - \dfrac{1}{2}g\sin{b} \cdot t^2$

    $y = u\sin(a-b) \cdot t - \dfrac{1}{2}g\cos{b} \cdot t^2$

    $y = 0 \implies t = \dfrac{2u\sin(a-b)}{g\cos{b}}$

    substituting for $t$ in the range equation, $x$, yields ...

    $x = \dfrac{u^2}{g\cos^2{b}}\bigg[\cos{b} \cdot 2\cos(a-b)\sin(a-b)-\sin{b} \cdot 2\sin^2(a-b)\bigg]$

    using double angle identities ...

    $x = \dfrac{u^2}{g\cos{b}}\bigg[\cos{b}\sin[2(a-b)]-\sin{b}+\sin{b}\cos[2(a-b)]\bigg]$

    $\dfrac{dx}{da} = \dfrac{2u^2}{g\cos{b}}\bigg[\cos{b}\cos[2(a-b)]-\sin{b}\sin[2(a-b)]\bigg]$

    using the sum identity for cosine ...

    $\dfrac{dx}{da} = \dfrac{2u^2}{g\cos{b}} \cdot \cos[b +2(a-b)] = \dfrac{2u^2}{g\cos{b}} \cdot \cos(2a-b)$

    $\dfrac{dx}{da}=0 \implies \cos(2a-b)=0 \implies 2a-b = 90^\circ \implies a = 45^\circ + \dfrac{b}{2}$
    Attached Thumbnails Attached Thumbnails max range of projectile-projectile2.jpg  
    Thanks from markosheehan
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  6. #6
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    Re: max range of projectile

    I am having trouble understanding some of the double angle identities.
    i can not see how 2sin^2 (a-b) goes to sinb*cos[2(a-b)]

    i also can not see how u^2/gcosb [cosb*sin2(a-b) -sinb*+sinbcos2(a-b) goes to
    2u^2/gcosb [cosb*cos2(a-b) -sinb*sin2(a-b)
    i can not see how you are factorizing by 2

    in essence i can understand lines 7-10 inclusive of your answer
    thanks
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  7. #7
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    Re: max range of projectile

    $\cos (2\theta) = 1-2\sin^2 \theta$
    $2\sin^2(a-b) = 1-\cos (2(a-b))$

    So, $x = \dfrac{u^2}{g\cos b}\left(\cos b \sin(2(a-b)) - \sin b + \sin b \cos(2(a-b))\right)$

    Then, skeeter took the derivative:

    $\dfrac{dx}{da} = \dfrac{u^2}{g\cos b}\left[\cos b (2)\cos(2(a-b)) - 0 + \sin b (-2)\sin(2(a-b))\right]$

    skeeter simplified:

    $\dfrac{dx}{da} = \dfrac{2u^2}{g\cos b}\left[\cos b \cos(2(a-b)) - \sin b \sin(2(a-b))\right]$
    Thanks from skeeter
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