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Math Help - Assistance needed with Projectile motion and max range

  1. #1
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    Assistance needed with Projectile motion and max range

    Hello,
    I was wondering if I could get some help with a class work problem. I'm in no way asking for just the answer. I need help understanding what the problem is asking for and a push in the right direction

    The problem reads:
    For the general projectile of exercise 31 ( see below), with h=0 (a) show that the horizontal range is
    Assistance needed with Projectile motion and max range-eq1.png
    and (b) find the angle that produces the maximum horizontal range.

    1. I not really sure how to how that the equation is in fact the range.
    2. I think that i need to calculate the directive of the equation and set it equal to 0 in order to find the maximum. Problem is Im not sure what to differentiate. It feels like its missing a part of the equation, time or distance maybe?

    Thank you for any help. Again Not looking for the strait answer just some help


    Exercise 31:

    Beginning with newton's second law of motion, derive the equations of motion for a projectile fired from altitude h above the ground at an angle (Theta) to the horizontal and with the initial speed v(sub 0)
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  2. #2
    Junior Member beebe's Avatar
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    Re: Assistance needed with Projectile motion and max range

    1. This is how I would do it, but it is probably not the fastest or easiest way.

    Start with y-y_0=(v_0sin(\theta))t-\frac{gt^2}{2}
    find t when y=0.

    Then go to x-x_0=(v_0cos(\theta))t with your value for t and simplify. You'll need a trig identity here. Notice that x-x_0=\Delta x=R.

    2. You already have  R=\frac{v_0^2sin(2\theta)}{g} . Think of this equation as the function R(\theta). How do I find the maximum values of that function?
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  3. #3
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    Re: Assistance needed with Projectile motion and max range

    Ok so my first question is where did you get T from?

    When I solve for t how ever I get t=(2*sin(Theta)*v)/g and t=0. I'm having a hard timing picturing this in my head.

    what to I insert as T in the x-x0=... and what do I use for x?

    As for the 2nd part I'm pretty sure I find the derivative of it... d/d(theta) = (2cos(2*theta)*v^2)/g right? buts that the max high correct, how do I find where its maximum x value is?

    PS. how are you making equation in the post like that? I had to use math type and upload it.
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  4. #4
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    Re: Assistance needed with Projectile motion and max range

    t = \frac{2v_0 \sin{\theta}}{g}

    substitute for t in the \Delta x equation ...

    \Delta x = v_0 \cos{\theta} \cdot t = v_0 \cos{\theta} \cdot \frac{2v_0 \sin{\theta}}{g} = \frac{v_0^2 \cdot 2\sin{\theta}\cos{\theta}}{g}

    now ... remember your double angle identity for sine?
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  5. #5
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    Re: Assistance needed with Projectile motion and max range

    Yeah should be (v^2*sin2x)/g right?
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  6. #6
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    Re: Assistance needed with Projectile motion and max range

    Quote Originally Posted by Dieseltwitch View Post
    Yeah should be (v^2*sin2x)/g right?
    \theta instead of x, but you get the idea.
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  7. #7
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    Re: Assistance needed with Projectile motion and max range

    Awesome. However I have no idea what that does for me. I want to understand what its doing. what does the substituting t in do?
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  8. #8
    Junior Member beebe's Avatar
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    Re: Assistance needed with Projectile motion and max range

    Quote Originally Posted by Dieseltwitch View Post
    Ok so my first question is where did you get T from?

    When I solve for t how ever I get t=(2*sin(Theta)*v)/g and t=0. I'm having a hard timing picturing this in my head.
    This is correct. This equation can be thought of as a function that returns the time that a projectile travels for a given starting velocity and angle, if the starting and ending heights are the same. If you put in some angle and velocity, then t= how long the projectile was in the air.

    what to I insert as T in the x-x0=... and what do I use for x?
    See Skeeter's post.
    Quote Originally Posted by Dieseltwitch View Post
    Awesome. However I have no idea what that does for me. I want to understand what its doing. what does the substituting t in do?
    x-x_0=(v_0cos(\theta))t

    The range is the change in horizontal position, so in this equation, x-x_0 is the range. v_0cos(\theta) is the part that gives you the initial horizontal velocity v_0x. Then that's multiplied times time. Hopefully, you can see how this equation is a form of displacement=velocity*time. Remember that we are ultimately working towards an equation where range is in terms of velocity and angle. Well, we just found an equation a little bit ago with time in terms of velocity and angle, so we just substitute. That gets rid of the t in the range equation and all that's left is thetas, velocities, and constants. All you have to do from there is algebra to get it to look like what it's supposed to.

    As for the 2nd part I'm pretty sure I find the derivative of it... d/d(theta) = (2cos(2*theta)*v^2)/g right? buts that the max high correct, how do I find where its maximum x value is?
    let range=R=R(\theta)

    The derivative you found is \frac{dR}{d\theta}. Recall that the derivative of f(x) lets you find the values of x that maximize or minimize f(x).

    PS. how are you making equation in the post like that? I had to use math type and upload it.
    LaTeX. See this thread.
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