



    // TODO Touch Up Event


        Fruit_BD.type = BodyType.DynamicBody;
        Fruit_BD.position.set(x, y);

        Fruit_FD.density = 1.0f;
        Fruit_FD.friction = 0.7f;
        Fruit_FD.restitution = 0.2f;

        MassData mass = new MassData();
        mass.mass = 5f;

        Fruit_Body[n] = world.createBody(Fruit_BD);

        System.out.println("Eggs... " + n);


        is_Next_Fruit_Touched = false;
        up = y;

        if(n < 50)




    return true;


    camera_1 = new OrthographicCamera();
    camera_1.viewportHeight = 1280;
    camera_1.viewportWidth = 720;



It's basic physics: heavier objects don't fall faster!


drag_force = object_specific_constant * velocity^2




  • 2
    \$\begingroup\$ I believe this is the most correct answer. Use different drag co-efficients for the different objects to achieve different fall rates. Most physics engines support this e.g Box2D. Although I suggest you revise your answer to include what Majd has stated that Mass does not typically affect fall speed. \$\endgroup\$ – Stephen Tierney Nov 26 '12 at 10:36
  • \$\begingroup\$ Let's see what I can do for your answer. \$\endgroup\$ – David Dimalanta Nov 26 '12 at 13:10
  • \$\begingroup\$ Although everything about this answer is correct, adding a drag force would be the incorrect way to handle the problem. Box2D already has built-in functionality for this. Not to mention, the equation you provide is pretty sketch. \$\endgroup\$ – Amplify91 Nov 26 '12 at 17:01
  • \$\begingroup\$ @Amplify91 What's sketch about it? It's the actual formula for air resistance. \$\endgroup\$ – Orin MacGregor Nov 26 '12 at 18:03
  • 2
    \$\begingroup\$ I can see how it is confusing, I didn't even understand the name at first. It's just a constant that is specific to that object. In a real world scenario it would primarily be (and simplified to) the surface area being acted upon by the wind resistance. For programming it would just be an arbitrary number that would cause the object to behave as intended. Ccxvii should probably put this as an explanation in the answer. \$\endgroup\$ – Orin MacGregor Nov 26 '12 at 20:46


    Don't be afraid to set that value well above 1.



  • \$\begingroup\$ Also, make sure your world's gravity is high enough to make your objects fall as fast as you want them to. Real world gravity is -9.8, but you may want to go higher. \$\endgroup\$ – Amplify91 Nov 26 '12 at 17:07
  • 1
    \$\begingroup\$ About number 1, setLinearDampling(), is it possible to set a negative number, over -9000, for even heavier? \$\endgroup\$ – David Dimalanta Nov 27 '12 at 2:00
  • \$\begingroup\$ @DavidDimalanta Can't tell if troll... I've set it around 10 before. \$\endgroup\$ – Amplify91 Nov 27 '12 at 2:12


  1. Set the velocity of the body after creation

    Body.setLinearVelocity(Vector2 v)

  2. Apply a force to the body after creation Body.

    Body.applyForce(Vector2 force, Vector2 point)

  3. Apply an impulse to the body after creation

    applyLinearImpulse(Vector2 impulse, Vector2 point)

  • 1
  • 1
    \$\begingroup\$ I can use Majd's idea for the next project later on. But for now, my current project, a stack-up game for the Android, needs to "cheat" the speed fall since this game tests the reflexes and puzzle solving skills. Best of all, kid friendly. \$\endgroup\$ – David Dimalanta Nov 27 '12 at 1:38


Wish you luck.

  • 1
    \$\begingroup\$ Where can I find it to adjust the pixels per meter thing? \$\endgroup\$ – David Dimalanta Nov 27 '12 at 7:37
  • \$\begingroup\$ When you creating new box2d object you just need to devide all the parameters by your devider, for example 30. I mean the following: \$\endgroup\$ – Mikle Pankratov Nov 28 '12 at 6:18
  • 1
    \$\begingroup\$ No. You can play with bodys mass, density, gravity, but what I said is the most common bug. \$\endgroup\$ – Mikle Pankratov Nov 28 '12 at 11:31
  • 1
    \$\begingroup\$ yeap, he is right. quote: "you should convert the values received from Box2D from meters to pixels. This will improve the stability of the physics simulation." link: box2d.org/2011/12/pixels \$\endgroup\$ – Mikle Pankratov Nov 28 '12 at 13:16
  • 1
    \$\begingroup\$ =) good luck!!! \$\endgroup\$ – Mikle Pankratov Dec 3 '12 at 10:48

Just increase the gravity acceleration, there is a basic formula:

V = 0.5 * g * t^2

g -> gravity acceleration t -> time in the air

  • 3
    \$\begingroup\$ I don't think this adequately answers the question, considering there is another already established and accepted answer. \$\endgroup\$ – kevintodisco Oct 16 '13 at 16:23

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