How Do You Solve the 2D Ball Bounce Problem with Angle and Velocity Constraints?

[Gil-galad]

Hello! Since this is my first post I would like to say I am happy that I joined PF and I will try to contribute as much as I can to this great community.

So let's get down to the gist of the problem.

You are standing 4m from a vertical wall and 0.8m from the ground. You have a bouncy ball that you throw with 10m/s initial velocity at some angle $$5^{0} \leq \alpha \leq 45^{0}$$ from the horizontal towards the ground. The ball bounces from the ground then bounces off the wall losing 10% of its initial velocity after each bounce. The question is to find this initial angle $$\; \alpha$$with less than $$0.01^0$$ of an error, so that the ball returns to the thrower (original position) assuming that the ball bounces off the wall/ground with the same angle as it hits it (incident angle=reflected angle). Assume gravity at 9.81 m/s/s and no friction is involved. The ball does not rotate.

So my question is how do you approach such problem? Is it possible to construct an analytical expression and then perhaps maximize/minimize only with the help of trivial kinematic equations? Or one must use some generalized coordinates and langrangian & hamiltonian mechanics? I would appreciate some guidance.

I pondered quite for some time before my calculations became quite complicated and nearly gave up. I used the trivial kinematic equations for the trajectory of the ball and tried to get an expression involving the initial angle related to the subsequent reflected ones that define the other 2 trajectories but to no avail.

 

[tiny-tim]

Welcome to PF!

Hello Gil-galad! Welcome to PF!

(have an alpha: α and a theta: θ and a ≤ )

You are standing 4m from a vertical wall and 0.8m from the ground. You have a bouncy ball that you throw with 10m/s initial velocity at some angle $$5^{0} \leq \alpha \leq 45^{0}$$ from the horizontal towards the ground. The ball bounces from the ground then bounces off the wall losing 10% of its initial velocity after each bounce. The question is to find this initial angle $$\; \alpha$$with less than $$0.01^0$$ of an error, so that the ball returns to the thrower (original position) assuming that the ball bounces off the wall/ground with the same angle as it hits it (incident angle=reflected angle). Assume gravity at 9.81 m/s/s and no friction is involved. The ball does not rotate.

So my question is how do you approach such problem? …

(i think it means "losing 10% of its initial speed" )

Just call the angle θ, and calculate the speed and angle before and after each bounce …

what do you get?

 

[nlightner]

Hello! It's great to have you as a new member of the PF community. This is definitely an interesting problem to tackle.

To approach this problem, we can use the principle of conservation of energy and momentum. Since there is no friction and the ball does not rotate, we can assume that the total energy and momentum of the system remain constant throughout the bounces.

First, let's define the variables we will be using:

- \alpha: initial angle of the ball
- v_0: initial velocity of the ball
- h: height of the ball at the top of its bounce (after hitting the wall)
- d: distance of the ball from the wall after the first bounce (when it returns to its original position)

Using the principle of conservation of energy, we can write the following equation:

\frac{1}{2}mv_0^2 = mgh + \frac{1}{2}mv_0^2 (0.9)^2 + \frac{1}{2}mv_0^2 (0.9)^4 + \frac{1}{2}mv_0^2 (0.9)^6 + ...

Where m is the mass of the ball and g is the acceleration due to gravity.

We can simplify this equation to:

v_0^2 = 2gh + \frac{v_0^2}{10}(1 + (0.9)^2 + (0.9)^4 + (0.9)^6 + ...)

Note that this is a geometric series, so we can use the formula for the sum of an infinite geometric series to simplify it further:

v_0^2 = 2gh + \frac{v_0^2}{10} \left( \frac{1}{1-0.9^2} \right)

Solving for h, we get:

h = \frac{v_0^2}{20g} \left( \frac{1}{1-0.9^2} - 1 \right)

Next, using the principle of conservation of momentum, we can write the following equation:

mv_0 \cos \alpha = mv_0 \cos \alpha (0.9) + mv_0 \cos \alpha (0.9)^3 + mv_0 \cos \alpha (0.9)^5 + ...

Simplifying this equation, we get:

\cos \alpha