Here is a program that has a ball drop and bounce from wherever the mouse is clicked. Would anyone know how to change the rate that the ball drops to the force of gravity?
Im trying to figure out the proper solution to this… but i am having a little trouble. All help and or input would be much appreciated.
float x;
float y;
float yspeed = 0;
float xspeed = 0;
float balldiameter = 10;
float ballradius = balldiameter/2;
void setup() {
size (400,400);
background (255);
fill (0);
ellipseMode(CENTER);
smooth();
noStroke();
x = width/2;
y = height/2;
}
void draw() {
mouseChecks();
boundaryChecks();
ballFunctions();
keyFunctions();
}
void mouseChecks() {
if (mousePressed == true) {
x = mouseX;
y = mouseY;
yspeed = mouseY - pmouseY;
xspeed = mouseX - pmouseX;
}
}
void boundaryChecks() {
if (y >= height - ballradius) {
y = height - ballradius;
yspeed = -yspeed/1.15;
}
if (y <= ballradius) {
y = ballradius;
yspeed = -yspeed/1.35;
}
if (x >= width -ballradius) {
x = width -ballradius;
xspeed = -xspeed/1.10;
}
if (x <= ballradius) {
x = ballradius;
xspeed = -xspeed/1.10;
}
}
void ballFunctions() {
if (balldiameter < 2) {
balldiameter = 2;
}
if (balldiameter > 400) {
balldiameter = 400;
}
ballradius = balldiameter/2;
background(255); //should this be in here?
ellipse (x,y,balldiameter,balldiameter);
yspeed = yspeed += 0.2;
xspeed = xspeed/1.005;
y = y + yspeed;
x = x + xspeed;
}
void keyFunctions() {
if (keyPressed) {
if(keyCode == UP) {
balldiameter +=1;
}
if (keyCode == DOWN) {
balldiameter -=1;
}
}
}
The acceleration of gravity on earth is 9.81 m/s^2. So, if your ball’s velocity is 0 at the point the mouse is clicked, the final velocity will be the acceleration integrated in relation to time. This would be (9.81 * t) / 2. Where t is in seconds. The resulting units would be m/sec. You would have to convert meters to some screen space unit for drawing.