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Throwing a Javelin (Projectile Motion)

  1. Oct 26, 2007 #1

    An athlete can throw a javelin 60m from a standing position. If he can run 100m at constant velocity in 10s, how far could he hope to throw the javelin while running? Neglect air resistance and the height of the thrower in the interest of simplicity.

    2. Relevant equations

    I've found the range to be R(θ) = 2 v0/g sinθ (v0 cosθ + vr)
    where v0 is the initial velocity and vr is the velocity of the thrower's initial run.

    3. The attempt at a solution

    Given the above equation and the premise that the thrower can throw 60m from standing, I solved for v0 using vr=0, R=60, g=9.8, and θ=45° as that is the ideal throwing angle from standing. I found v0 = √(60g).

    Next I tried to find the ideal angle to throw at when vr = 10 m/s. I did this by taking solving R'(θ)=0 where R'(θ) is the derivative of the range formula. I found:
    R'(θ) = c v0²/2 cos2θ + c vr cosθ = 0 where c = 2 v0/g.

    This yielded the equation v0 cos2θ = -vr cosθ. Using an identity for cos2θ, I obtained the following quadratic equation:
    2 cos²θ + d cosθ -1 = 0 where d = vr/v0 = 10/√(60g).

    Solving for θ I got θ = 0.65 rads = 37°. The book, however, lists 52.3° as the angle, and therefore gets a different value of R than I do. I'm not sure where I went wrong with my logic.
    Last edited: Oct 26, 2007
  2. jcsd
  3. Oct 26, 2007 #2


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    I haven't look through the detail yet, but assuming that the runner/thrower throws at the same angle as when standing, the runner's velocity adds to the vx of the javelin, but does not contribute to the vertical velocity.

    Make sure the correct angle is calculated for the range (60 m) of the throw from stationary position.
  4. Oct 26, 2007 #3
    That's kind of what's tying me up. The problem stated nothing about the initial angle thrown, so I made the assumption that the thrower threw at the angle to maximize range. From standstill that's pi/4, and with some added vr, I presumed the ideal angle would change and sought to calculate it and use it to calculate the final range.
  5. Oct 26, 2007 #4


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  6. Nov 3, 2007 #5
    Okay, so after thinking about this, I've realized the angle clearly should be greater than [tex]45^\circ[/tex]. This is because the [tex]v_x[/tex] is greater than the [tex]v_y[/tex] when the thrower is running. That being the case, he must throw at a greater angle so that the effective angle thrown at is [tex]45^\circ[/tex]. Provided that reasoning correct, I have, for maximum range:

    [tex]v_{0x} = v_{oy}[/tex]
    [tex]v_0 \sin\theta = v_0 \cos\theta + v_r[/tex]
    [tex]\sin\theta - \cos\theta = \frac{v_r}{v_0}[/tex]

    Squaring both sides gives

    [tex] \sin^2\theta + \cos^2\theta - 2\sin\theta\cos\theta = \frac{v_r^2}{v_0^2}[/tex]
    [tex] \sin2\theta = -\frac{v^2_r}{v^2_0}[/tex]

    However, when I solve that I don't get the angle of [tex]52.3^\circ[/tex] (I get something ludicrous like [tex]-10^\circ[/tex]). Was my reasoning incorrect?

    Also, when I take

    [tex]R = \frac{2v_0}{g}(v_0 \cos\theta + v_r) \sin\theta[/tex]

    and solve R' = 0 I get [tex]35^\circ[/tex] as the solution. Where am I going wrong here?
    Last edited: Nov 3, 2007