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Projectile Motion and impact.

  1. Sep 4, 2007 #1
    1. The problem statement, all variables and given/known data
    A gun on the shore (at sea level) fires a shot at a ship which is heading directly toward the gun at a speed of 40 km/h. At the instant of firing, the distance to the ship is 15,000 m. The muzzle velocity of the shot is 700 m/s. Pretend that there is no air resistance.
    (a) What is the required elevation angle for the gun? Assume g = 9.80 m/s^2.
    (b) What is the time interval between firing and impact?

    2. Relevant equations

    Uhmm, I'm guessing. tflight = [tex]\frac{2v_{0} sin\alpha }{g} [/tex]

    Maybe xmax = [tex] \frac{v^{2}_{0}sin2\alpha}{g} [/tex] will also be applicable.

    [tex] v_{x} = v_{0x} = v_{0}cos\alpha [/tex]
    [tex] v_{z} = v_{0z} - gt = v_{0} sin \alpha - gt [/tex]
    [tex] x = v_{0x}t [/tex]
    [tex] z = v_{0z}t - \frac{1}{2} gt^2 [/tex]

    3. The attempt at a solution
    I'm not sure what to do with this problem. Any way I set it up, I end up with more variables than I can solve for.

    I would appreciate a hint to throw me in the right direction. Thanks in advance.
  2. jcsd
  3. Sep 4, 2007 #2


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    You seem to be throwing in equation willy-nilly. Draw a diagram, and do the usual thing of splitting up into horizontal and vertical components, and use the kinematic equations. Your horizontal target will be moving, so x_max=1500-40t (original distance minus the distance the ship covers in the time the bullet is in the air). You should be able to write an equation for the x component of the displacement of the bullet. Eliminate t using an equation for the y component of the displacement of the bullet.
  4. Sep 4, 2007 #3
    [tex] 40 \frac{km}{h} * 1000 \frac{m}{km} * \frac{1}{3600} \frac{h}{s} = 11.1 \frac{m}{s} [/tex]

    [tex] x_{max} = 15,000 - 11.1 t [/tex]

    [tex] x(t) = 700 cos \alpha t [/tex]

    [tex] 700 cos \alpha t = 15000 - 11.1 t [/tex]

    [tex] z(t) = \frac{-1}{2} (9.80) t^2 + 700sin \alpha t [/tex]

    I don't know what to do now.. But I'm pretty sure my equations are good.
  5. Sep 4, 2007 #4


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    Ok, well use the fact that the displacement, z, is equal to zero when the bullet hits the ship. Can you then solve this equation for t? (Hint: quadratic formula)
  6. Sep 4, 2007 #5
    [tex] 0 = \frac{-1}{2} (9.80) t^2 + 700 sin \alpha t [/tex]

    By the quadratic equation,

    [tex] t = 142.86 sin \alpha [/tex]

    So plugging t in.. I get..
    [tex] 100000 sin \alpha cos \alpha = 15000 - 1587.302 sin \alpha[/tex]

    How do I solve this for [tex] \alpha [/tex] ?
    Last edited: Sep 4, 2007
  7. Oct 2, 2011 #6
    Use the 2 graph method, I just got this problem for homework myself. The angle should come out to around 8.6, and the time to around 21.4 seconds. Hope this helps
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