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Projectile motion: solving for variable angle

  1. Jan 10, 2012 #1

    I've been muddling over this problem for a few days. I thought there would be a simple approach, but I'm having trouble reaching a solution. Here is the rundown:

    We have a 2D projectile launcher. It hurls an object from the origin (0,0) at some initial velocity with magnitude [itex]V_i[/itex] at angle θ with respect to the ground. You will be given its initial velocity, and must choose the launch angle θ such that the projectile will hit a target at point x,H (that is, a point x distance away with height H) during its descent (i.e. anytime after the peak where [itex]V_y = 0[/itex] ). It is a reasonable assumption that θ will be in the range [itex] 0 \leq θ \leq \frac{\pi}{2} [/itex].

    I am trying to find an explicit formula for θ in terms of [itex]V_i [/itex] , [itex] x [/itex] , [itex] H [/itex], and the usual [itex] g = -9.8 \frac{m}{s^2} [/itex]
    [itex] V_{yi} = sin \theta V_i [/itex]
    [itex] V_{xi} = cos \theta V_i [/itex]

    We have the following solutions for t:
    [itex] V_{yi} t + \frac{1}{2} g t^2 -H = 0 [/itex]
    [itex] V_{xi} t - x = 0 [/itex]

    Since we want the projectile to hit the point (x,H) at the same time, we set the equations equal to each other:

    [itex] \frac {1}{2}g t^2 + ( V_{yi} - V_{xi} ) t + (x - H) = 0 [/itex]

    Use the quadratic formula to find t. (I expanded [itex] V_{xi} [/itex] and [itex] V_{yi} [/itex] in the resulting formula so we have θ back in the equation ).

    [itex] t = \frac{-V_i ( sin \theta - cos \theta ) \pm \sqrt{{V_i}^2 ( 1 - sin{2 \theta} ) - 2 g (x - H)}}{g} [/itex]

    I used trig substitution in the square-root to make a [itex] sin^2 + cos^2 = 1 [/itex] and double angle formula for [itex] 2 sin( \theta ) cos ( \theta ) = sin( 2 \theta ) [/itex] .

    I want to factor that nasty expression under the radical by 'completing the square', but I don't see how that is possible.

    Also, I don't really want to solve for t ... that is, I don't care how long the projectile takes. All I care about is what angle it needs to be launched at so that it hits point (x,H) .
    Last edited: Jan 10, 2012
  2. jcsd
  3. Jan 10, 2012 #2
    If you don't need t, then why solve for it?
    You can eliminate t between the two equations and find an equation in the variable theta.
    For example, find t from the second one and plug it into the first one.
    Then you can write the cos in terms of tan and find a quadratic equation in tan(theta).
  4. Jan 11, 2012 #3
    Aha... yes, I didn't really want to solve for t, I just didn't see a way around it. Your approach is simple, and I'm dumbfounded I didn't realize it first ;)

    However, it still seems to be an involved problem. Now we have:

    [itex]t = \frac{x}{V_0} [/itex]
    Plugging this t into the formula for [itex]V_y[/itex], I get:
    [itex] tan ( \theta ) x + \frac{ g x^2 }{2 cos^2 ( \theta ) V_0^2 } - H = 0 [/itex]

    This is very close to an explicit formula for θ, but I can't find any trig identities that will help me simplify. I can convert the [itex] \frac{1}{cos^2}[/itex] into [itex]sec^2[/itex] , but... that doesn't really seem to help.
  5. Jan 11, 2012 #4
  6. Jan 11, 2012 #5
    Hm, I still had problems finding the solution. I 'cheated' and used WolframAlpha to solve that last formula for theta.

    Wolfram uses substitution and gives:

    [itex] \theta = tan^{-1} \left ( \frac{ -V_i^2 \pm \sqrt{-g^2 x^2 + 2 g H V_i^2 + V_i^4}}{gx} \right ) [/itex]

    WolframAlpha then states that after plugging the substitution back into the formula and running a check, it determines this solution is 'incorrect'. I plugged in some test numbers a few times, and the value for theta appears correct (consistent).
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