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Angled Projectile at an elevated height w/o a given angle

  1. Sep 24, 2012 #1
    1. The problem statement, all variables and given/known data

    Many modern 3D video games must be designed such that objects move in ways as close as possible to reality. This is often called “game physics”, which is run by a physics engine coded into the game. In one event of a recent game, a player’s job is to shoot a cannon at objects on a distant cliff. The target is supposed to be at a distance of 400 m and is at a slight elevation with respect to the player – say 150 m – and the launch speed of the cannonball is supposed to be 100 m/s relative to the player's “stationary” avatar. The game physics ignores wind resistance (many if not all of them do). At what angles should the cannon be fired in order to hit the target?


    Enter the larger angle in degrees.

    What is the smaller angle?

    What is the time difference between the two trajectories?

    ay=-9.81 m/s^2
    dx=400 m
    dy=150 m
    Vo= 100 m/s
    Angle?
    time?

    2. Relevant equations

    Vf=Vo+at
    d=Vot+(1/2)at^2
    d=(Vf^2-Vo^2)/2a

    3. The attempt at a solution

    I found the greater angle to be 77.6 degrees. And by experience usually the smaller angle is simply 90-whatever the angle is to find the other. The smaller angle measures to be 12.4 degrees but this can't be right due tot he fact that the angle of elevation of the object is 20.55 degrees so the angle must be at least that. I'm not sure where to progress.

    I made up a general formula for finding the distance (found by v=d/t)

    d=(10000sin(2angle))/9.81

    Help!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 24, 2012 #2

    SammyS

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    Using 90° - Angle1 = Angle2 only works if the target is at the same elevation as the launcher (cannon).

    How did you find the first angle?

    You gave few details as to how you arrived at your answer.

    Added in Edit:

    Hint: The time it takes for the projectile to reach the target is the same as each of the following:

    The time it takes an object to travel 400 m at a constant velocity equal to the horizontal component of the velocity of your projectile.

    The time it takes an object tossed straight up to return to an elevation of 150 meters if it is tossed upward with an initial velocity equal to the vertical component of the velocity of your projectile.
     
    Last edited: Sep 24, 2012
  4. Sep 24, 2012 #3
    I found my answer by putting 150^2+400^2=d^2 so that I could put d into my equation that i showed earlier.I came out with a value for theta, and it was 12.4, and the question asked for the larger angle so i substracted it from 90 degrees to get 77.6
     
  5. Sep 24, 2012 #4

    SammyS

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    12.4° can't be right. The line of sight, from the cannon to the target makes an angle of about 20.56° above the horizontal.
     
  6. Sep 25, 2012 #5

    SammyS

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    This is one of the harder type of projectile problem to do --- outside of those with air resistance.

    Write out the kinematic equations for horizontal motion. This is constant velocity, so these are relatively easy.

    Write out the kinematic equations for vertical motion. This is constant acceleration, not so simple as the above.

    When I combine these I get either a quadratic equation in tan(θ) or a quadratic equation in t2 . Both look quite messy, although I left them in terms of parameters & variables, without substituting many numbers.
     
  7. Sep 25, 2012 #6
    When setting up the quadratic formula for tan(θ), do you let the vertical component equal 150?
     
  8. Sep 26, 2012 #7

    SammyS

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    The vertical component of what?

    What is your equation?
     
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