What are all the equations we use to calculate how bounces work?

Question

I mean, what is the object's final displacement, or the function that describes the object's height over time (see [1]) of an object thrown by a height $h$ with a speed of $\vec{v_0}$, a mass of $m$, a "bounce factor" of $\Lambda$ and the floor with a friction of $F_f$ and the air with a friction of $F_a$.

[1]

Answer 1

I've also answered a similar question here.

Variables

I'm using the subscript $y$ to denote stuff in the perpendicular direction (along the $y$ or $h$ axis), and $x$ for stuff in the parallel direction (along $x$).

I'll use $u$ for initial velocities and $v$ for final velocities. The initial and final refer to "just before/after a bounce", and "just before/after an arc", where "bounce" refers to the moment when it touches the ground, and "arc" is the arcing motion afterwards.

$e$ is the coefficient of restitution--this is the "bounciness" you wanted. It can have any value from 0 to 1, where 0 is completely unbouncy(inelastic), and 1 is very bouncy. (elastic). A value greater than one gives an unphysical effect where it bounces higher. It's related to energy via $\text{loss in energy} = \frac12mv^2(1-e^2)$, for a ball bouncing on the ground. So I guess $\Lambda_\text{(percentage of energy loss)}=1-e^2\times100\%$

$\mu$ is the friction coefficient for the ground-ball interface. It is 0 for a frictionless surface, and usually less than 1. Not necessarily, though. It will lead to an additional loss of energy not taken into account by $e$ or $\Lambda$.

$g$ is gravitational acceleration

Note that all the quantities are signed.

Relevant formulae, condensed version

For a bounce

$$v_y=-eu_y$$ $$v_x=u_x+ \mu(e-1)u_y$$

These $v$s become $u$s for the upcoming arc.

Bounces are pretty much instantaneous. If you want to consider the time factor, you need to know the shape and young's modulus of the object.

For an arc

The arc will be executed in a time $t_{arc}=2u_y/g$, and will attain a maximum height $y_{max}=\frac{u^2}{2g}$

During this time: ($t$ is the time since the arc started, NOT the total time) $$y=u_y-\frac12gt^2$$ $$x=u_xt$$

At the moment the arc finishes,

$$v_x=u_x$$ $$v_y=-u_y$$ These $v$s become the $u$s for the next "bounce"

All the forumulae

SUVAT equations

See suvat-equations. In this case, $a_y=-g,a_x=0$, and $s$ is distance travelled in relevant direction. You can apply these equation separately for $x$ and $y$.

Bouncing

Here, $N$ is normal force. $f$ is the friction force. $J$ refers to impulse, $p$ to momentum. Friction $$f=\pm\mu N \text{ direction can vary}$$ Impulse $$J_y=\int N\rm dt$$ $$J_x=\int f\rm dt$$ We can combine these to get $J_x=\pm\mu J_y$ We take the sign $-$ in this case, as friction opposes motion, and the motion in $x$ direction is positive.

$$J=\Delta p=m(v-u) \text{ for both axes}$$

Combining all these, you can get the bounciness equations.

Note: if you want your ball to have a spin as well, the equations become more complicated.

Answer 2

I would use a combination of projectile motion and restitution/collision equations for a simple model. Model each bounce individually using projectile motion eqs. for trajectory and for each bounce collision, use restitution/collision equations to calculate the angle of launch for the next bounce as well as initial energy/velocity.