The Bounce Shot

Now I am going to show you how to humiliate your opponent. The opponents first cup has already been drank and there are nine left. If you skimmed over the rules section you may have missed the bounce shot. If the bounce shot is accomplished, the opponent must drink two cups, but he may also attempt to swat the ball out of the way. First, shift yourself perpendicular to the table and line your shot parallel to your opponent's cup. The distance between you and the middle cup should be 2.3 meters. You must still throw at a speed of 6 m/s, this won't be hard unless you have already had to drink a few cups. Now we must find the angle at which we need to toss. This simple answer lies within some simple equations.

We will use The Range equation to calculate the first distance (see the graph above) is d1 = ((v_o² * sin(2θ))/g). This equation is derived by eliminating t and combining vectors.

Now d2 is derived from the horizontal motion equation: (v_o*cosθ)t

However, we also need the time, which can be solved for using the vertical motion equation, h=v*t+1/2 * at² (Halliday & Resnick).

Since theta is the angle desired we must change the velocity to equal v_o * (sinθ), and by setting this equation equal to zero and acceleration equal to g we can finish this by writing 0 = v_o*(sinθ)*t+1/2*gt² –h

By using the quadratic formula the only possible equation without a negative time is Time = 1/g (-v_o*sinθ + (v_o²*sin²θ + 2gh)^1/2)

So after substitution into the d2 equation the distance between d1 & d3 becomes d2 = (v_o*cosθ)* 1/g (-v_o*sinθ + (v_o²*sin²θ + 2gh)^1/2).

Now the last distance equation seems difficult but is just simple equations we have used previously. Since theta is the desired variable, we must solve for theta again. By using tangent we are able to find the angle at which the ball is coming in from, but we need to break it down into X & Y components. So v_x = v_ocosθ, the common vector equation. Since gravity acts upon the velocity in the y direction we can form the equation v_y = v_osinθ + g*t, and we already have the equation for t, so just substitute it in. This allows us to somewhat find θ, now θ = tan-1(v_osinθ + ((-v_o*sinθ + (v_o²*sin²θ + 2gh)^1/2)))/ v_ocosθ, after the time substitution is included then gravity is eliminated. To calculate the range we just use the d1 equation, ((v_o² * sin(2θ))/g). But now we must take into effect the energy lost in the bounce. If we use the gravitational potential energy equation (mass * gravity * change in height), we can find the potential energy lost. But better yet, let’s just use the ratio between the two! After numerous tests on a ping pong table I used the final height divided by the initial height to prove that a loss of 22% occurs after the first bounce. So we use the range equation and multiply µ into it, which is equivalent to 78%. So our equation becomes ((µ*v_o² * sin(2θ))/g). Now we also have an equation for theta that we solved for previously, so let’s shove that in there also.

Now the last distance d3= ((µ*v_o² * sin(2* tan^-1(v_osinθ + ((-v_o*sinθ + (v_o²*sin²θ + 2gh)^1/2)))/ v_ocosθ))/g).
Awesome, we have all three equations now. We now need to add them all up and solve for the total distance, D = d1 + d2 + d3:

D = ((v_o² * sin(2θ))/g) + (v_o*cosθ)* 1/g (-v_o*sinθ + (v_o²*sin²θ + 2gh)^1/2) + ((µ*v_o² * sin(2* tan^-1(v_osinθ + ((-v_o*sinθ + (v_o²*sin²θ + 2gh)^1/2)))/ v_ocosθ))/g)

So depending on your height, table dimensions, speed, gravity, and theta, it can all be calculated with this equation. This equation is very similar to the golden ratio in that it has an endless amount of possibilities + beer, so I like to refer to it as the Platinum Equation.

We will now substitute all of our values into this Platinum Equation to find out which values of theta will work best:

2.3 = ((6² *sin(2θ))/9.8) + (6*cosθ)*1/9.8 (-6*sinθ + (6²*sin²θ + 2*9.8*.88)^1/2) + ((0.78*6² *sin(2*tan^-1(6*sinθ + ((-6*sinθ + (6²*sin²θ + 2*9.8*.88)^1/2)))/ 6*cosθ))/9.8)
θ = 79.66° & -64°

I am not going to bust out my mad trig skills, but instead I will resort to using the solve function on my ti89 to solve for theta. After waiting about 8 minutes for the calculator to solve for the results we finally end up with θ = 79.66° & -64°. This shows that if you toss the ball at an angle of 79.6° to the horizon it will bounce and land straight into the opponents’ cup. Also, if you throw it downwards at an angle of -64 to the horizon, it will land into the cup. I set the height equal to 0.88 meters because this is the height after the height of the cup is subtracted. Sir Isaac Newton will now demonstrate the results of our calculations.

An alternative to this problem allows us to find a target to aim for. If we add the equations d1 and d2 we can find the distance in which the ball needs to travel in order to bounce successfully. If we substitute 79.5° into theta we are able to produce a distance of 1.46 meters.

I will not go into great detail outlining the rest of the shots. If a cup is 4” to the left, then just move 4” to the left. If it is a greater distance, then sub your own values into the Platinum Equation and you will have the correct angle. The rest of the game is common sense, but depending on how much you have consumed, your accuracy will decline similar to -e^x does on a coordinate plane.