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I am not 100% sure, but I think that by resetting the ball to the bottom line when it actually arrives a few pixels below the bottom line you exactly compensate the elasticity factor, so that the ball bounces on indefinitely. Try removing the line <code>yPosition=(displayHeight-20);</code>, then it should work better. This way, the ball will eventually come to a rest, but this is still not correct, as the time the ball falls through the ground is not counted at all. Update Okay, I think now I've got it. Here are the relevant bits: <pre><code>// first update position, then update velocity yPosition=yVelocity+yPosition; if (yPosition >= floor) { // invert velocity and position w/ elasticity applied yVelocity = -yVelocity * elasticity; yPosition = floor - (yPosition - floor) * elasticity; } if(Math.abs(yVelocity) <= 2 && Math.abs(yPosition - floor) <= 2){ // stop everything when close to rest yPosition=floor; yVelocity=0; } else { // otherwise accelerate, even after bounce yVelocity=yVelocity+yAccelleration; } </code></pre> So the main changes w.r.t. your code are: <ol> <li>first to update the position, and then the velocity</li> <li>instead of resetting the ball to the floor, reverting the amount it fell through the floor, and also applying the elasticity factor to that</li> <li>always to accelerate the ball, even when it bounces off the floor</li> </ol> And here how it looks like (floor being at 500). Not perfect, but close. <img src="https://i.stack.imgur.com/mjfBO.png" alt="enter image description here"> Update 2: Just found <a href="http://processing.org/learning/topics/bouncybubbles.html" rel="nofollow noreferrer">this Processing example</a>, which is actually very close to your original code, with the difference that they always apply acceleration, and that they do so before the collision-checks. Seems like this alone does the trick.
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1. COI did not mean to remove the if-check, but the line below that. (so, basically the same as Roberto suggests; did not see that he edited his answer). However, instead of removing the line it might be better to invert the fall-through and multiply it with the factor... working on that.
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2. COThank you :D Three thumbs up
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