Investigating the physics of basketball Research Paper

Investigating the physics of basketball - Research Paper Example Concepts of momentum and collision are helpful in studying the behavior of ball when it hits the ground. On the other hand, equation of air resistance can be used to establish the favorable atmospheric conditions for playing basketball. Gravity and Projectile Motion A projectile is an object that has been thrown into space and is moving under the influence of gravity (Nag, Pati and Jana 16). The path of movement of a projectile is referred to as trajectory (McLester and Pierre 282). When the object does not encounter any force apart from gravity, its path is a parabolic (Goswami 28). A basketball in motion is an example of a projectile. This is because it moves under the influence of gravity once it is thrown by the player. In order to model a basketball motion as projectile motion some assumptions must be made. First, the effect of air resistance is ignored. Second, the ball is treated as a particle. Third the acceleration due to gravity is taken to be the same. Fourth, the spinning motion of the ball is assumed to be minimal. Under these conditions, the equations of motions can be applied to determine various motion parameters of the basketball. When a player throws a ball, it moves up to a maximum height, Ho. The speed of the ball decreases and at Ho the speed becomes zero. The ball then falls back with an increasing speed. The basketball player ought to know that the angle at which he/she throws the ball will affect the horizontal displacement of the ball. The horizontal displacement in this case is the distance the player is from the ring mast. The player must also know the right thrust to give the ball in order for it to reach the ring. Consequently, the kinematic equations are very helpful to the player. When a projectile is thrown to space at an angle, its velocity at any given point has two components; the horizontal component and the vertical components. The horizontal velocity is constant while the vertical velocity changes because of acceleration du e to gravity. If a projectile is launched with initial velocity Vo at an angle ? from the horizontal, its initial vertical velocity is Vo Sin ? while its initial horizontal velocity is Vo Sin ?. The horizontal displacement is given by the equation x = Vo Cos ? t .This equation resembles the formula for getting distance for one dimension motions. The horizontal velocity is constant for vertically launched projectiles. The equation is used to find the range of the projectile. The vertical displacement of projectile is given by the equation S = ut + 1/2 gt2 where u is initial velocity g is acceleration due to gravity and t is the time. For projectiles launched at an angle ?, the equation becomes; S = Vo Sin ? t - 1/2 g t2 †¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦.. (1) Making t the subject; T = x /Vo Cos ? t †¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦ (2) Replacing S= Vo Sin ? (x / Vo Cos ?) -1/2 g (x /Vo Cos ?) 2........................... (3) Simplifying (3) S= x Tan ? – x2 (g / 2Vo2 Cos2 ?) †¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦ (4) A player throwing a ball is at distance R from the ring mast. The height of the ring mast is H1 and the height of the player is H2 as shown in the diagram below. The vertical displacement of the ball from where it was thrown is given by; Y = x Tan ? – x2 (g / 2Vo2 Cos2 ?) †¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦ (5) Y defines the displacement of t