II. The Problem, Constants, Variables, and AssumptionsIt is noticeable that players can often make errors while shooting, but the shot still goes in. I’veplayed varsity basketball before, and based on my experience I assume that the angle of release is themain reason that allows for the amount of error a player can make in their free throws. Often times I couldshoot with a little bit too much strength or in the wrong direction, but because I’d shot at a good angle thefree throw had a lot more room for error. As such, the model will be based on trying to find the best anglea basketball player of 7’1 (Shaq) can shoot a free throw.The physical constants that play a role in governing a free throw are as follows:D r : The diameter of the rim, 1.5 feetD b : The diameter of the ball, 0.8 feetL: Horizontal distance traversed, 13.54 feet (rather than using 14 feet, the exact distance from thefree throw line to the center of the hoop, researchers in the past (Hamilton, G.R. and Reinschmidt,C.) have made mention that players release the ball a little bit in front of the line, so we thus takeL as 13.54 feet rather than an exact 14 feet.)h: Vertical distance traversed, 1.146 feet (it has been observed by researchers ( x) that players, onaverage, release the ball at 1.25 times the shooters own height. As we are initially modellingShaq, a 7’1 player, we then take 1.146 feet as vertical distance traversed.)g: Acceleration due to gravity, -32 ft/s -2 (9.8 m/s -2 converted to feet)2Figure 2.1: Court DimensionsThe assumptions made in creating the model are as follows:1. Ignore air resistance. Air resistance, while adding much complexity to the model, has a veryminor effect on it.2. Ignore ball spin. Spin is unimportant as we assume the ball only goes into the net, or hits theback of the rim before going in.3. Use trajectories that either go directly into the net or trajectories that hit the back of the rim andthen go in, so no sideways error in the trajectory. This is done to cover a wide range of successfultrajectories while also retaining simplicity for the model creation.4. No error in initial shooting velocity. The first model only focuses on shooting angle, rather thanshooting velocity.5. As we are using Shaq, the shooter is 7’1. After finding Shaq’s best angle, this assumption canbe removed and the model will be applied to shooters of different heights.3Figure 2.2: Free Throw Diagram (G, Joerg andIII. Model DerivationFig 3.3: Initial VelocityIn a free throw, the basketball is thrown at a certain angle from the ground ( ) and ? its trajectoryin flight can be seen as parabolic as a result of Isaac Newton’s laws. As a parabola, the flight path of thebasketball goes both vertically and horizontally. As such, I will begin with Newton’s projectile motionequations at an initial velocity (v 0 ), separating it into horizontal and vertical components:4(1) V h = V 0cos?and(2) V v = V 0sin?respectively, ? 0 being the initial release angle. The subscript 0 is used as conventionally it refers to aninitial value of a variable, where in this case the ball has not yet moved and is at an initial value.Horizontally, there is no air resistance as per the assumptions made previously, so the horizontal equationof motion is:(3) x(t) = V t ,where x(t) is distance, V is velocity, and t is time. I should substitute the initial horizontal velocityequation into this function as my objective is to solve for a model that looks for the initial angle, so I get(4) x(t)= v0tcos(?)Using L as the horizontal distance traversed, and T as the time it takes for the ball to travel to the basket,x(T) = L can be substituted into the last equation with the objective of obtaining an equation that I canderive to find T, which will be incorporated into the equation for vertical velocity. Gravity, as thisequation only accounts for horizontal distance traversed, will only be included for the equation of verticalmotion.(5) L = v0Tcos(?)A similar process can be applied for the vertical equation of motion, which is given by (6)y(t) = vt + 2gt , where g is acceleration due to gravity (included as the function acts vertically). I can1 2then substitute vt for the initial vertical velocity equation for the same reasons as earlier, obtaining(7) y(t) = v0sint(?)+ 2gt1 2Again simply substituting y(T) for h, or vertical distance traversed, I obtain the following:(8) h = v0sinT(?) + gT 21 25As I mentioned I would do earlier, from (5), we can take the value of T as (9) T = L , thenv0cos(?)substituting this solution into (8) and finally solving for the initial velocity:(10) h = v0sin(?) g ,Lv0cos?+ 21 Lv0cos?2Which I further simplified and derived to obtain an initial model for initial velocity,(11) v0 = sin(?) L ,cos(?)? ?g2(Ltan(?)?h)although the model does have physical restrictions in that the ball is limited to forward motion (0 < ? 0 <90?), as the objective of a free throw is to move the ball forward. What this initial model allows us to dois to find the initial velocity at which the ball passes through the center of the rim at any release angle ?0 ,so for example if I released a free throw at an angle of 49 °, I would have to shoot with an initial velocityof ? 22.7m/s. At this point, I realized that I would have to find a way to gauge the space for error atdifferent release angles, which I wasn't sure about. So I consulted studies such as those by Gablonsky andLang who opted to work out the horizontal position of the ball as it comes down to the basket height.They do this by deriving the equation for amount of error through keeping initial velocity fixed andallowing the release angle to vary. As I understand it, and going back to what I said about angle ofrelease, a great angle of release can negate some of the errors in someone's free throw form, so theydecide to keep the initial velocity at a fixed value to observe how the amount of error changes based on achanging release angle. to As such, the equation will be derived to calculate the length of error byreplacing with and introducing , referring to a less accurate release angle than the ideal L x ?e0?0where the ball is assumed to go through the middle of the ring:(12) x = (v sin(?) ) ?gv0cos(? )e00 +?v0 sin(?) gh2+ 26IV. InequalitiesFrom there, I turned again to Gablonsky and Lang who derive some criteria for the basketball tostill go in the net, one being to avoid contact with the front of the rim, and the other having the ball hit theback of the rim. I assume that to avoid contact with the front of the rim, the distance between the center ofthe ball and the rim must have remained larger than the ball's radius ( ) throughout the 2 entire motion.DbIt can also be noted that x + 2 is the horizontal distance of the rightmost part of the ball when the ballDbis level with the basket. Taking the distance between the ball's center and the rim as s , the followingcriterion was set by the researchers:(13) s x(t) (L )) y(t) ) ) , = ( ? ? 2Dr + ( ? h > ( 2Dbwhich I interpret as a limit that defines how the ball avoids contact with the rim, as o alwaysremains greater than the radius of the ball: they never meet.(14) x + , 2Db = L + 2Drwhich I interpret as the only situation where the amount of error allows for the ball to touch therim at a straight horizontal trajectory, as the ball’s radius, when at the end of the shooting length, is levelwith the rim, meaning that the ball always bounces into the hoop considering the lack of spin.To finally solve the equations and find an answer for the error allowed for Shaq’s initial angleknown as ? 0 , v0 is kept fixed and the boundaries ?low < ? < ?high are set to solve the assumption that s isequal to the radius of the ball (a shot that does not touch the rim), s ? , as well as the assumption 2Db= 0that the ball hits the back of the rim before going in, x ? L + . 2Db?Dr = 0I learned in class that to find the maximum of a function, one is to differentiate it and then findthe zero of the derivative, as the derivative's maximum should amount to 0. I ran through the process andtried to solve e'( ?0) = 0, but found that the function was not differentiable. I again decided to consult7Gablonsky and Lang as this was a problem that I hadn't ever encountered in the past. They shared similarthoughts, mentioning that they had to use a specific computer algebra system's optimization routine to getthe best release angle of ? ? 48.18? requiring the release velocity of v ? 21.7 ft/s. I wanted to downloadtheir MATLAB code and take a look at how they found the maximum point numerically, but the linkwritten in the footnote had died.We can now assume that Shaq, at 7'1, has his best free throw release angle at 48.18?, althoughthis is assuming that he is able to keep his release velocity at 21.7 ft/s.V. Conclusion and Final ThoughtsThe assumption that Shaq (or any other player, for that matter) would be able to keep his releasevelocity consistent between every attempt is completely unrealistic, and was only used for the purpose ofmaking the model simpler to solve. As such, the model will have to be refined further in order to takeinto account the consistency of a player's release velocity. In an ideal model, no assumptions should bemade about the situation in order for the model to perfectly emulate its real world counterpart, anachievement that I clearly do not fulfill in my exploration. For instance; I do not factor in air resistance,ball spin, nor do I look at different flight paths or trajectories. I'd describe my model as very rudimentary,and one that has potential to include many other factors that affect a free throw. In addition, there aresome factors that couldn't be worked into a mathematical model such as, for example, the psychologicaleffects of shooting a free throw in front of a crowd. While the model is still missing a lot, it is stillimportant to note that it does have merit in solving what it was meant to do: finding the room for error atdifferent free throw release angles. In conclusion, the model, while unable to completely reflect on thereal-world factors that go into a free throw, it still succeeds in modelling the length of error for differentrelease angles of a free throw, albeit unrealistically.