Any child can pick up a basketball, bounce it a few times, and throw it at the hoop without realizing the complex physics he is involving himself in. However, a greater understanding of the sport can be gained by applying the principles of physics to these actions. Through this paper I will be examining the physics of dribbling and shooting.

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For the equations, I will use a standard men’s basketball with a mass of 25 oz. (0.7 kg) and a diameter of 9.39. Dribbling If the basketball was simply dropped from a height of 1 meter with no initial push from the dribbler and rebounded from the floor in a perfectly elastic collision, Mechanical Energy for the system would be conserved. However, the collision is not completely elastic and the dribbler pushes the ball with an initial force to ensure that it returns to his hand. To simplify the process I will first consider a situation in which the ball is allowed to fall freely from a height of 1 meter and has a perfectly elastic collision with the floor. After that I will take the true nature of the collision with the floor and the force applied by the dribbler into consideration.

Because Mechanical Energy (ME) is equal to Kinetic Energy (KE) plus Gravitational Potential Energy (GPE) and Mechanical Energy is conserved, I know that the initial KE + GPE at 1 meter equals the KE + GPE just before it impacts the ground. If KE = ?? mv2 and initial velocity is zero, then the initial Kinetic Energy is zero. GPE would equal mass X gravity X height. Assuming mass is 0.7 kg, gravity is 9.8 m/s2 , and the height is 1 meter, then GPE is equal to 7.0 kg * m2/s2. Conversely, the GPE would be zero just before impact because height is zero but KE would be .5(.7 kg)(v2). Knowing that initial ME equals ME just prior to collision, we can determine that velocity equals 4.5 m/s. If the ball rebounds with perfect elasticity, then it would return to its original height of 1 meter. We know from observation that the basketball does not have a perfectly elastic collision because it does not return to its original height after impact. The elasticity depends on the air pressure inside the ball and the surface the ball is bounced on. If the ball is not adequately inflated, then the collision time with the floor will increase, causing a decrease in the elasticity. According to Louis A. Bloomfield, Professor of Physics at the University of Virginia, on his website titled How Things Work, in which he discusses the physics of common activities, The more pressure a basketball has inside it, the less its surface dents during a bounce and the more of its original energy it stores in the compressed air. That stored energy is transferred into the balls return bounce. If pressure decreases, some energy is transferred into heating the bending skin of the ball and less energy is conserved to be applied to the return bounce. An over-inflated ball would have too great of an elastic collision. Although dribbling would be easier, shooting would be more difficult because the ball would bounce too far off of the backboard and rim. Spalding, a leading maker of basketballs, recommends inflating a basketball to 7-9 pounds per square inch for optimal performance. Most basketball courts are made of solid wood or concrete, which will not dent substantially from the force of the basketball. However, if the ball is dribbled on a surface like grass, the grass will absorb part of the energy and turn it into heat. A ball dribbled on this surface would not bounce as high as one dribbled on concrete. Because the collision is not elastic and it is desirable to have a faster dribble to prevent opponents from stealing the ball, the dribbler applies a downward force on the ball.

For a fast dribble this force is more than enough to return the ball to its original height. In fact, the dribbler slows the balls ascent by pulling his hand up as the ball begins to make contact with it, increasing the collision time so that the ball does not simply bounce off of his hand. Increasing the collision time also decreases the acceleration of the ball because acceleration equal to the change in velocity divided by the time taken for the change. By decreasing the acceleration, the force of the ball on the hand is also decreased because force is equal to mass x acceleration. This is easily observed by trying to catch a fast moving ball. If a person’s hand is held stationary, they will feel a sting from the impact of the ball with their hand. However, if they move their hand with the ball to increase the collision time, they will not feel the sting. Shooting Shooting a basketball involves propelling it from an initial height of about 2 meters, depending on the height of the basketball player, to the rim of the basket at 3.05 meters and at a varying distance. Because projectile motion can be divided into its vertical and horizontal components for simplicity, I will examine the vertical component of shooting using conservation of Mechanical Energy and a few other formulas. By setting the point of reference for the shot at its resting point in the hand of the shooter, the initial GPE is equal to zero because height is zero. When considering the vertical component of the shot, the arc must be taken into account. In his article Underhanded Achievement in Discover Magazine, Curtis Rist advocates increasing the arc on the ball’s trajectory because it gives the ball a better chance of going in. If the arc is too shallow, there is only a small margin between the diameter of the ball and the rim because the rim appears as an ellipse from the ball’s perspective. As the arc increases, the rim takes on a more circular shape, making an easier target for the ball because of the enlarged diameter.

Therefore, the ball must be given a sufficient push to give it an initial velocity capable propelling it to a height of about 4 meters so that it will be able to fall down into the hoop. Given that the GPE is zero at rest in the shooters hand, the only relevant component of the ball’s Mechanical Energy just after it leaves the shooter’s hand is its Kinetic Energy. The KE is equal to ?? mv2, where the mass again is 0.7 kg. By setting this equal to the ball’s ME at the top of its arc, we can determine the initial velocity of the ball required for it to reach a height of 4 meters. At its peak, the only relevant component is GPE because its kinetic energy is zero. Setting KEinitial equal to GPEpeak we have 0.5(.7kg)(v2) = (.7kg)(9.8m/s2)(4m). By working out the equation, we know that the initial velocity must be 6.5 m/s. If we estimate the motion of the shot to take about 0.3 seconds, we can calculate the acceleration and force involved. Acceleration equals change in velocity divided by change in time, giving an acceleration of 22 m/s2. Plugging this into the equation for force (Force = mass x acceleration), we can determine that the force necessary to propel the ball in the vertical direction is 15.2 N. After the ball leaves the hand of the shooter, the constant downward force of gravity is the only force that acts on it (negating wind resistance). When the ball reaches it maximum height the force of gravity pulls it down to the hoop at 3.05 meters. If it falls about 1 meter, it will reach a vertical velocity of 4.4 meters per second just before it goes through the rim. Although we cannot determine the force and velocity required for the horizontal component of shooting because they vary depending on the distance to the hoop, we can make some general statements about the horizontal component. After the ball leaves the shooter’s hand, the only force acting in the horizontal direction is wind resistance. Wind resistance should be minimal unless the basketball player is shooting outside on a windy day. According to Newton’s first law of motion that states, Every object remains at rest or in motion in a straight line at a constant speed unless acted on by an unbalanced force, the object will continue at a constant velocity until it comes in contact with another force. Another aspect of shooting is the rotation of the ball. The shooter gives the ball a backward rotation to cause it to bounce down when it hits the backboard or rim. The rotation gives the ball rotational inertia and angular momentum.

Rotational inertia is the property of an object that measures its resistance to a change in its rotational speed. It remains constant for the basketball because the diameter of the ball remains constant at 9.39 for a standard men’s ball. Comparatively, a standard women’s basketball with a diameter of 9.07 would have a smaller rotational inertia because of its smaller diameter. Angular momentum is equal to its rotational inertia multiplied by its rotational speed. The rotational speed will vary depending on the shooter. Applying the right hand rule to the backward rotation of the ball, we can describe the angular momentum as a vector that points to the right from the center of the ball (as the ball heads away from the viewer). The backspin given the basketball has another interesting effect. As the ball spins, the dimples grab the air around the ball and pull it through the spin, creating a layer of air spinning with the ball. The airflow underneath the ball has a higher speed compared to the air around it than the air flowing over the top of the ball. This causes the air underneath the ball to break up faster than on top of the ball, shifting the wake downward. The momentum exerted to shift the wake down exerts an upward momentum on the ball, causing the ball to drop more slowly than it would with no spin. Therefore, giving the ball backspin decreases the force necessary to propel a basketball a certain distance because it will not fall as quickly. Understanding the physics of shooting and dribbling a basketball may not increase the enjoyment of watching the NBA finals, but it may help improve a basketball player’s skill. If he can find the right force necessary to propel the ball back to his hand when dribbling and realizes the importance of giving the ball a chance to fall down into the rim, he will be better off for the next game.

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