Bouncing ball

Relevant wiki: Conservation of Energy

The maximal potential energy after each bounce is reduced to $81 \,\text{\%}$ , so that the height $h_n$ is reduces by the same factor: $\begin{aligned} E_{\text{pot},n} &= m g h_n \\ E_{\text{pot},n} &= 0.81 \cdot E_{\text{pot},n-1} \\ \Rightarrow \quad h_{n} &= 0.81\cdot h_{n-1} = 0.81^n \cdot h_0 \end{aligned}$ The equations of motion for the ball reads $\begin{aligned} x(t) &= x_n + v_x t \\ y(t) &= v_{y,n} t - \frac{1}{2} g t^2 \end{aligned}$ with the time $t$ after the last bounce and the initial velocity $v_{y,n}$ . We solve for the maximum point $y(t_\text{max}) = h_n$ by zeroing the derivative: $\begin{aligned} \frac{d}{dt} y(t_\text{max}) &= v_{y,n} - g t_\text{max} \stackrel{!}{=} 0 & \Rightarrow \quad t_\text{max} &= \frac{v_{y,n}}{g} \\ y(t_\text{max}) &= \frac{v_{y,n}^2}{2 g} = h_n & \Rightarrow \quad v_{y_n} &= \sqrt{2 g h_n} \end{aligned}$ Therefore, we can estimate the position of the next bounce: $\begin{aligned} x_{n+1} &= x_n + v_x \cdot (2 t_\text{max}) = x_n + 2 v_x \sqrt{\frac{2 h_n}{g}} \\ x_\infty &= v_x \sqrt{\frac{2 h_0}{g}} + \sum_{n = 1}^\infty 2 v_x \sqrt{\frac{2 h_n}{g}} = v_x \sqrt{\frac{2 h_0}{g}} \left[1 + 2 \cdot \sum_{n = 1}^\infty (0.9)^{n} \right] \\ & = v_x \sqrt{\frac{2 h_0}{g}} \left[1 + 2 \cdot \left(\frac{1}{1 - 0.9} - 0.9 \right) \right] = 2.6 \,\frac{\text{m}}{\text{s}} \cdot 0.4 \,\text{s} \cdot 19.2 = 20 \,\text{m} \end{aligned}$ Here, we have used the geometric series $\sum_{n = 0}^\infty q^n = \frac{1}{1 - q}$ In other words, the bouncing ball is the symbolization of the geometric series. Although the ball jumps up almost infinitely often on the ground, the times between two jumps become shorter, so that the ball comes to rest after finite time and distance.

Relevant wiki: 2D Kinematics Problem-solving

Time of first bounce is $t_{0}=\sqrt{\frac{2h_{0}}{g}}=0.4s$ .

Kinetic energy is directly proportional to the square of velocity, so the velocity after a bounce is $v_{n}=\sqrt{1-0.19}v_{n-1}=0.9v_{n-1}$ . The time spent between the bounce and the highest point is directly proportional to velocity (as $v=gt$ ), so the time spent between points $x_{1}$ and $x_{2}$ is $t_{1}=2*0.9t_{0}=0.72s$ (multiplication by two is necessary in the first case, as the ball starts from the top position before the first bounce.

Every other $t_{n}=0.9t_{n-1}$

$t_{1}, t_{2}, ...$ is a convergent infinite geometric series, the sum is $\frac{t_{1}}{1-0.9}=10t_{1}=7.2s$

So the bouncing ends after $0.4+7.2=7.6s$ , while the ball travels $7.6*2.6=19.76m$ horizontal distance.

The ball will roll after this - it doesn't stop, as there are no horizontal forces.

Relevant wiki: 2D Kinematics Problem-solving

My idea is to concentrate on the vertical motion of the ball, and find the total time it is in the air. Once this is found, multiplying it by the given horizontal component of the velocity (2.6) will then give the distance travelled.

Let us work out the total air time for the extended problem in which the trajectory begins from ground level and includes the other half of the first parabolic arc. In other words the ball starts a distance $x_1$ to the left of the y axis. of course the flight time for this extended problem will be greater than for the actual problem, and we will make an adjustment later.

Suppose the vertical component of the launch velocity is $V$ . The time taken to reach the peak on the y-axis is $\frac{V}{g}$ and so the flight time for the first parabola is $\frac{2V}{g}$ .

For each subsequent parabolic bounce, the vertical component of the launch velocity is reduced by a factor of $\sqrt{1-0.19}=0.9$ . Since the flight times are directly proportional to the vertical component of the launch velocities, we can combine this with the previous result to get the total flight time for the extended problem to be

$\frac{2V}{g}(1+0.9+0.9^2+0.9^3+......)$

Summing the infinite series and setting g = 10, the extended flight time is thus 2V.

Now we can subtract the flight time for half of the first parabola to find the flight time for the actual problem to be

$2V-\frac{V}{g}$

Setting g=10 this gives a flight time of $1.9V$

Now we just need the initial upward velociy. This comes from the conservation of energy

$\frac{V^2}{2}=gh$ and so with g = 10 and h = 0.8 we find $V=4$

And so the flight time is $4 \times 1.9$

Multiplying this by the horizontal velocity gives the total bounce distance as

$4 \times 1.9 \times 2.6=\boxed{19.76}$