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Math Help - Find theta given

  1. #1
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    Cool Find theta given

    Given the distance a projectile needs to travel, the initial velocity, and acceleration due to gravity, how can I calculate the launch angle?
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    I'm assuming \Delta y = 0.

    \Delta y = v_{0y}t - \frac{1}{2}gt^2

    0 = v_{0y}t - \frac{1}{2}gt^2

    0 = t(v_{0y} - \frac{1}{2}gt)

    t = \frac{2v_{0y}}{g}

    \Delta x = v_{x} t

    \Delta x = v_{x} \cdot \frac{2v_{0y}}{g}

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

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

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

    \theta = \frac{1}{2}\arcsin\left(\frac{g \Delta x}{v_0^2}\right)
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by quaz579 View Post
    Given the distance a projectile needs to travel, the initial velocity, and acceleration due to gravity, how can I calculate the launch angle?
    let the launch angle be \theta. now, i suppose we are launching from the ground.

    you should know that V_{x_0} = v_0 \cos \theta and v_{y_0} = v_0 \sin \theta

    where v_0 is the initial velocity, v_{x_0} is the horizontal component of the initial velocity and v_{y_0} is the vertical component of the initial velocity. we obtained these equations by using trig ratios on the various direction vectors.

    now, work in feet, since you didn't specify. so acceleration due to gravity is -32 feet per (second squared). also, s(t) is the default for the position function, v(t) is the default for the velocity function and a(t) is the default for the acceleration function.

    thus, dealing with the vertical component:

    a_y(t) = -32

    \Rightarrow v_y(t) = \int a_y(t) ~dt = -32t + C

    since v_y(0) = v_{y_0} = C we have

    v_y(t) = -32t + v_{y_0}

    \Rightarrow s_y(t) = \int v_y(t)~dt = -16t^2 + v_{y_0}t + D

    assuming the initial height is zero, clearly s_y(0) = D = 0

    so, \boxed{s_y(t) = -16t^2 + v_{y_0}t} ----------- this is a formula that gives us the height at time t.


    dealing with the horizontal component:
    a_x(t) = 0

    \Rightarrow v_x(t) = \int a_x(t)~dt = C

    since, v_x(0) = v_{x_0} = C, we have

    v_x(t) = v_{x_0}

    \Rightarrow s_x(t) = \int v_x(t)~dt = v_{x_0}t + D

    since s_x(0) = 0 = D we have

    \boxed{s_x(t) = v_{x_0}t} ---------- this is a formula for the horizontal distance traveled.

    now, all of that is standard procedure, you should practice it a lot and thus be able to do all that pretty quickly with minimum thinking (always a good idea on an exam ), but here's getting to the problem

    if d is the horizontal distance traveled, then we wan

    s_x(t) = d

    that means v_{x_0}t = d

    \Rightarrow t = \frac d{v_{x_0}}

    \Rightarrow \boxed{t = \frac d{v_0 \cos \theta}}


    also, for this same t, we want s_y(t) = 0 (since we want the object to hit the ground at this time. (of course if we are not going to the ground, just set s_y(t) = \text{whatever}.

    so, s_y(t) = -16t^2 + v_{y_0}t = 0

    \Rightarrow t (-16t + v_{y_0}) = 0

    \Rightarrow t = 0 ..........this is when we start, we don't want this t

    or

    -16t + v_{y_0} = 0

    \Rightarrow t = \frac {v_{y_0}}{16}

    \Rightarrow \boxed{t = \frac {v_0 \sin \theta}{16}}


    now remember, these t's are the same, so we can equate them. thus, we have

    \frac {v_0 \sin \theta}{16} = \frac d{v_0 \cos \theta}

    \Rightarrow v_0^2 \sin \theta \cos \theta = 16d

    \Rightarrow \sin \theta \cos \theta = \frac {16d}{v_0^2}

    \Rightarrow \frac 12 \sin 2 \theta = \frac {16d}{v_0^2}

    \Rightarrow \sin 2 \theta = \frac {32d}{v_0^2}

    \Rightarrow \boxed{\theta = \frac 12 \sin^{-1} \left( \frac {32d}{v_0^2} \right)}

    there has to be an easier way to get to that formula. but i am not in a thinking mood now, obviously. i just followed a standard method


    EDIT: there, skeeter has the shorter way he ends up using formulas that i derived. i should have remembered those formulas!
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  4. #4
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    Thank you

    This isn't a huge deal. I actually don't need an easy way to get this formula just as long as it works. This is actually for a video game programming class which I need this formula for in order to write code that works properly. I took physics over 2 years ago and have forgotten A LOT.

    A quick question though: where did the -16 come from in step 2? Is this just (acceleration due to gravity)/2 and if so can I plug in 9.81/2 if i am using meters as my units?

    Thanks again
    Last edited by quaz579; September 29th 2008 at 05:32 PM. Reason: Figured something out
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by quaz579 View Post
    This isn't a huge deal. I actually don't need an easy way to get this formula just as long as it works. This is actually for a video game programming class which I need this formula for in order to write code that works properly. I took physics over 2 years ago and have forgotten A LOT.

    A quick question though: where did the -16 come from in step 2?

    Thanks again
    i would just use what skeeter did. they are standard physics formulas. i've never done physics, which is why i didn't remember them

    the -16 comes from integrating -32t. the integral of t is (t^2)/2, that 2 divided into the -32 to give -16

    anyway, if you are using a program, just program the formulas that skeeter gave. pretty all my boxed formulas but the last one is given by some physics formula. why re-invent the wheel?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by quaz579 View Post
    This isn't a huge deal. I actually don't need an easy way to get this formula just as long as it works. This is actually for a video game programming class which I need this formula for in order to write code that works properly. I took physics over 2 years ago and have forgotten A LOT.

    A quick question though: where did the -16 come from in step 2? Is this just (acceleration due to gravity)/2 and if so can I plug in 9.81/2 if i am using meters as my units?

    Thanks again
    if you are using meters, then you should start with -9.81 instead of -32. so my -16 would be your -4.9
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  7. #7
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    Thanks a million
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