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Classical projectile problem

  1. Apr 14, 2010 #1
    Hi I have a classical projectile problem:


    Car moving west at 50mph. If I am south ten miles and wish to his the object with a canon traveling at 500mph
    i) What bearing should I shoot at?
    ii) What pitch?



    Thanks
    Jo






    2. Relevant equations

    for projectile
    x=(Vo cos (Picth)) t
    y=(Vo sin(Pitch)t - 1/2 g (t*t)


    3. The attempt at a solution

    I got a bearing of 32 degrees (with North as 0 obviously) and a pitch 0f 47 degree.
    Is this correct?
     
  2. jcsd
  3. Apr 15, 2010 #2
    Actually I'm not sure the above is correct, heres my new thinking:

    If I use the fact that the x will be the same at collision
    Tank
    x = 50t

    Rocket P is the pitch
    x = 500Cos(p)t

    so 50t = 500Cos(p)t


    Also we can use the fact that on collision y = 0 for the rocket because its hitting the ground so

    y = 500Sin(p)t - 1/2g t*t = 0

    So we have 2 simultaneous equations and 2 variables.

    What do you think?
     
  4. Apr 15, 2010 #3
    I'm not sure what you mean in your question when you say the cannon is travelling at 500mph. Do you mean that this is the speed of the projectile fired from the cannon?
    Also, your measurements are in mph, so are you using g in m/h^2? It might be easier to convert into m/s for velocity, then use 9.81 m/s^2 for g.
     
  5. Apr 15, 2010 #4
    My above calculation is incorrect because the projectile motion X Y are in a different plane. The arc of motion of the cannonball. So I cant equate the X's.
    Yeah they were just given in mph and yes the cannonball speed is 500mph.
     
  6. Apr 15, 2010 #5
    ok, so when you use g for acceleration due to gravity you have to remember to use a figure in m/h^2.
    I must admit, I'm not an expert, and I've been thinking about this a bit - it's quite a complex question because there are two angles to consider. The trick must be to find the time of collision required to meet the car, then work back to find the angles required to create that time of landing.
    Sorry I can't help more
     
  7. Apr 15, 2010 #6
    Yeah I just covert both speeds to m/s
    Vcar = 22.352m/s
    Vcannon= 223.52m/s
     
  8. May 18, 2010 #7
    First of all, when talking about directions on the compass, it is standard to assume that due East is 0 degrees, due North is 90 degrees, due West is 180 degrees and due South is 270 degrees.

    I assume that the cannon is located 10 miles due south of the car at the instant that the cannon is fired. If that is correct, you have 2 similar right triangle vector diagrams (one based on distance, the other on speed). Since we already know that the projectile travels at 500 mph and the car at 50 mph, the speed vector diagram is the easiest to produce: In this case, the speed of the projectile is the hypotenuse of the triangle and the speed of the car is the shorter of the two perpendicular vectors (50 mph at 180 degrees). The vertical leg is then found by using the Pythagorean Theorem. This can be shown to be approximately 497.5 mph (at 90 degrees).

    Therefore, the bearing angle will be the angle represented by the 500 mph vector, which is slightly more than 90 degrees (and less than 135 degrees, since the angle between the 500 mph vector and the vertical vector is acute).

    Assuming the angle between the vertical vector and the projectile's vector is represented by [tex]\alpha[/tex]

    Trigonometry shows us that
    [tex]\alpha = sin^{-1} \left( \frac{50}{500} \right) \approx 5.74 degrees[/tex]

    Therefore the bearing would be 90 degrees + 5.74 degrees, or 95.74 degrees
     
  9. May 19, 2010 #8
    Looking back, this isn't quite right. This would be the angle if the horizontal component of the projectile's speed was 500 mph. Obviously, you would have to fire the projectile at some inclined angle above the horizon in order for it to reach its target. that means that the horizontal speed of the projectile is some value less than 500 mph. Therefore, the bearing angle would be even closer to 90 degrees (the offset angle would be less than 5.74 degrees).
     
  10. May 20, 2010 #9
    If I understand your problem correctly, you have a car that is traveling due west at 50 mph. You also have a cannon that can fire a projectile at 500 mph. You don't fire the cannon until the car is due north of you (which happens to be a distance of 10 miles).

    You want to know two things: 1) how many degrees you must lead the car when you fire the cannon, and 2) the azimuth (or inclination above the horizon) of the cannon at the time it is fired in order to hit the car.

    This is a trick question. At this distance, the projectile fired from the cannon cannot "catch up" to the car if it is traveling 50 mph. In fact, you couldn't hit the car if it wasn't moving (at a distance of 10 miles). A 500 mph projectile can only be fired a maximum distance of roughly 3.17 miles.

    All other things remaining the same, the projectile would have to be fired at minimum speed of slightly more than 889.77 mph in order to be able to hit the car.
     
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