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#CONSTRUCT 2 PHYSICS BALL BOUNCE TOUCH HOW TO#
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#CONSTRUCT 2 PHYSICS BALL BOUNCE TOUCH FREE#
Construct 2 Free is the fr Construct 2, free and safe download. Construct 2 Free latest version: Construct 2 Free - Create your own video games. Print("stopped bouncing at t=%.Construct 2 Free, free and safe download. T_last = -sqrt(2*h0/g) # time we would have launched to get to h0 at t=0 Hstop = 0.01 # stop when bounce is less than 1 cmįreefall = True # state: freefall or in contact Hmax = h0 # keep track of the maximum height To get from the above to a "mathematical model" you might want to modify this python code to your needs: from math import sqrt But for your simulation that is unlikely to matter. In reality, as the ball deforms more, the spring constant will increase - which will actually make the contact time shorter as the impact velocity is higher. While the ball is in contact with the ground, we can consider its center to move in an approximately sinusoidal fashion - as I said, the important thing is that the contact time is independent of the drop height (on the assumption of linear spring constant). The time spent in the air until the next bounce is given by $$t_1 = 2\frac$$
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Using $\rho=0.75$, if you drop a ball from a height of 5 m (and putting $g=10~m/s^2$ for convenience) it would land after 1 second with a velocity of 10 m/s, and after the bounce (some short time $\Delta t$) it would come up again with a lower velocity $v_1 = 7.5 m/s$. For a golf ball, this can be as high as 0.8 for a tennis ball it is around 0.75. The coefficient of restitution is a parameter of a ball/surface, and reflects the fraction of velocity just after the bounce divided by the velocity just before. As a simplifying assumption, consider the ball to be a "mass on a slightly lossy spring with a spring constant $k$" then the ball will exhibit (approximately) damped simple harmonic motion during the impact: this means that the impact time will be independent of the impact velocity(!) The exact shape of the ball during the impact is hard to capture in a simple equation because it depends on the composition of the ball and the floor. During the impact, the ball will deform and there will be friction.This velocity will change from one bounce to the next. While the ball is not in contact with the ground, the height at time $t$ after the last bounce at $t_0$ is given by $$h(t+t_0) = v_0 t - \frac12 g t^2$$ where $v_0$ is the velocity just after the bounce.That means we can break the problem into two parts: You say you are ignoring friction of the air, but assume a partially inelastic collision. I'm happy to continue reading about it until I understand, I just don't know where to turn at this point. I know there are constants to deal with: how much the ball resists compression and how much energy is lost during the bounce, but I don't know how they are related.Ĭan anybody give me a one- or two-paragraph explanation and/or (a) formula(e) and/or point me to a resource where I can read more? (yes, I've already spent an hour or so googling it) And I'm assuming negligible loss to air resistance.) The ball isn't traveling horizontally or rotating at all. (Also, I'm only dealing with strictly vertical linear velocity and acceleration. But I don't know how to model the compression and deflection of the ball hitting the ground. I know that it experiences some compression which translates it's velocity upward again, at which point gravity is again the only force acting on it. I fully understand the physics while the ball is rising and falling: It accelerates downward at 9.8 meters/second/second.īut once it hits the ground, I'm lost. I'm programming an animation of a bouncing ball, and I want it to be as realistic as possible.