Although there are several forces acting on a pickleball while it is flight, gravity is prevalent. It is considered to be the great equalizer because it acts on everything equally, and there is no way to avoid it – gravity always wins in the end.
This is the second article in a series describing Pickleball Dynamics. The reader is advised to read the previous articles to better understand this topic. It is useful to first study the field of Projectile Motion, as the basic concepts and equations will establish our foundation in our understanding of pickleball dynamics.
Projectile Motion Problem
Let’s first look at the field of projectile motion to bring out some basic concepts. Figure 1 shows a cannon capable of firing projectiles at different velocities from a cliff that is 100 ft above the ground. The cannon can fire only in the horizontal direction (θ=0) and the cannon operator (gunner) wishes to hit a target that is located 2000 ft from the cliff.
For the time being, we will ignore aerodynamic effects (drag, wind, and lift). The muzzle velocity (v0) is the speed of the projectile at the moment it exits the cannon. The elapsed time for the projectile to strike the ground (or time of flight) is denoted by T. Although the gunner knows that there is a linear relationship between the amount of charge (gunpowder) placed in the chamber vs distance traveled for the projectile, he does not know the exact amount of charge that is needed to hit the target. He will therefore attempt to establish this experimentally.
First, he places a small amount of charge into the chamber resulting in the projectile undershooting the target and traveling only 1000 ft (the red curve in Figure 2). On his next shot, he increases the amount of charge, resulting in the projectile overshooting the target and traveling 3500 ft (the blue curve in Figure 2). Knowing the linear relationship between charge and distance, on his third shot he puts the ideal amount of charge into the chamber and makes the perfect shot and hits the target at 2000 ft (the green curve in Figure 2).
Do you notice anything strange about the different projectile trajectories? Although the horizontal distance travelled varies greatly, the elapsed time or time of flight (T) for the projectile to hit the ground is always the same (2.48 sec)! The reason for this becomes obvious when we examine the so-called equations of motion of the projectile in the horizontal (x) and vertical (y) directions.
Equations of Motion
In the horizontal direction, the position of the projectile as a function of time, x(t), can be calculated by adding its initial position (x0) to the product of the muzzle velocity (v0) multiplied by the time (t):
x(t) = x0 + v0t Equation 1
In this case, the initial position (x0) equals 0. What is evident in Equation 1 is that the horizontal distance travelled is dependent only on the initial velocity of the projectile (v0). We cannot solve Equation 1 by itself for the initial velocity (v0) because we do not know the time of flight, T.
In the vertical direction, the muzzle velocity (v0) does not contribute to the vertical distance traveled, because the cannon is pointed in the horizontal direction. The projectile, however, is affected by gravity (g) and must accelerate towards the ground (negative y-direction) at a rate of -32.17 ft/s2 (-9.81 m/s2). So,
y(t) = y0 + ½gt2 Equation 2
In Equation 2, we know that the starting vertical position (y0) is 100′ and the ending vertical position (y at time T) must be zero when the projectile strikes the ground. Therefore, we can solve Equation 2 for the time the time of flight (T) it takes for the projectile to hit the ground, which turns out to be 2.48 seconds.
Knowing that T=2.48 seconds, we can substitute into Equation 1 and determine the required muzzle velocity (v0) of the projectile, knowing that the ending horizontal distance (x at time T) must be 2000’. This yields a result of 803 ft/sec as the required muzzle velocity to hit the target.
Increase Contact Height for Greater Speed
The big “take-away” here is that unless there is a downward component of velocity, the time it takes for an object (such as a projectile or pickleball) to hit the ground from a given height will be the same, regardless of how much horizontal distance the object must travel. Therefore, if you can contact the pickleball at a height that is higher than the top of the net you can hit it back harder and faster than a ball that you must contact below the height of the net. This is one of the reasons why good pickleball players return a looping return or lob in the air, as opposed to letting the ball bounce. This is an important lesson which applies to pickleball serves, dinks, and ground strokes.
In our article, “How Fast is a Pickleball Serve?” we apply the equations of motion to the game of pickleball to determine the theoretical maximum speed of serves. Later, we will expand our analysis to determine the effects of aerodynamic drag, wind, and lift (induced by spin), and apply the expanded equations of motion to serves, groundstrokes, dinks, and overhead smashes.
