Science behind Sepak

To optimize a serve in Sepak is to minimize the time of the ball in the air, the speed and manner of how it falls. This will cause difficulty for the opponents team harder for the to return the ball. To make it more clear, We'll illustrate a diagram.

S(a) represents the distance of the ball to be served by the server from the net

S(b) represents the distance to where the ball will land after service

c represents the angle of trajectory of the ball

Projectile Motion

           In Physics, air resistance is omitted when projectile motion is discussed. Thus, the trajectory of a projectile appears as a parabola in which the horizontal component remains constant while the vertical component is affected by gravity.   On the other hand, in sepak, both gravity and air resistance direct the ball’s trajectory.  This influence of air resistance leads to a smaller range than would in calculations neglecting it. In reality, the trajectory of the sepak is not as symmetric as a parabola.  Generally speaking, the higher the air pressure, the higher the air resistance. After the sepak is kicked, headed or served, the pressure of the air in front of it increases, causing air resistance to rise.  Furthermore, this rise in air resistance causes the sepak to be pushed back, slowing it down.  This is why the range is much shorter than it would be when neglecting air resistance.

g represents the acceleration due to gravity (9.8 m/s2)

ho represents the initial height of the ball after service

hmax represents the maximum height reached by the ball

H represents the height of net

Point 1 is the service location

Point 2 is the targeted landing of the ball

Headers and the use of Vectors

            When aiming a header, vectors must be taken into consideration.  According to physics, “the angle of incidence equals the angle of reflection.”  Therefore, to aim the header, one must head the ball in the direction from which it was served.  To obtain the final speed and direction of the sepak, the following vectors must be combined:  the motion of the head, the motion of the ball, and the bounce –reflection- off of the head.

Example:

A sepak player aims a header 2.00 meters above the ground at an angle of 32°.  If the initial velocity of the sepak is 8.00m/s, what is the distance traveled by the ball?  (Air resistance is negligible).

Solution:

Recall the various equations used in Projectile Motion problems.  We first need to determine the components of the initial velocity.

X velocity: Vx = V(cosθ)

Y velocity: Vy = V(sinθ)

V0x = 8.00m/s (cos32°) = 6.78m/s

V0y = 8.00m/s (sin32°) = 4.24m/s

Now we can use the y velocity to calculate the amount of time the sepak was in the air.  We can set y = 0 and y0 = 2.00 to show that we want to know when the ball hit the ground.  Recall:

Y = y0 + v0yt + (1/2)ayt2

Y = 2.00 + 4.24t + (1/2)(-9.8) t2 = 0

Thus, using the quadratic formula, t = 1.20s

To find the range, we can substitute this time value into the equation for the x position:

X = x0 + v0xt

X = 0 + 6.78(1.20) = 8.16m

Head impact power

                     From the general expression for the rate of change of transitional and rotational kinetic energy for any rigid object is given by:

Power = Linear force (Linear velocity) + Rotational Torque (Angular Velocity)

In this study, only linear impact is encountered. Thus the head impact power can be calculated as:

HIP = Impact Force (Linear Velocity)