Data Type  Value

Velocity (x)

4.026 m/s

Velocity (y)

5.905 m/s

Velocity (xy)

8.662
m/s

Momentum (x)

2.508
(Kg)(m)/(s)

Momentum (y)

3.679
(Kg)(m)/(s)

Momentum (xy)

5.396 (Kg)(m)/(s)

Acceleration (x)

0 m/s^{2} ^{}

Acceleration (y)

10.353 m/s^{2} ^{}

Acceleration (xy)

10.353 m/s^{2}

Force (x)

0 N

Force (y)

6.457 N

Force (xy)

6.457 N

Impulse (x)

.0997 N/s

Impulse (y)

0.255 N/s

Impulse (xy)

0.998 N/s

Work (x)

0 N*m

Work (y)

3.668 N*m

Work (xy)

5.595 N*m

Inspired by
the YouTube of Kobe Bryant jumping an Aston Martin, our group decided to choose
a basketball related topic finally choosing the physics of a jump shot. Building upon the knowledge of the concepts
of velocity, acceleration, mass, force, momentum, impulse, work, and the
conservation of energy we learned in class we were able to devise the plan of
measuring those unknowns required to properly analyze the physics of a jump
shot. The tools used to measure the
variables were a laptop, lab pro, wii mote as an accelerometer, and a force
plate, scale pole, and the high speed camera.
The first unknown we chose, as well
as found necessary to find was velocity.
To do this we relied on the high speed video and tracker, with a known
distance; from the scale pole and a known time we were able to then determine
the velocity in the x, y, and xy coordinate planes. We found that in the x direction the ball
traveled at a speed of 4.026 m/s and in the y 5.905 m/s. The high velocity in the y direction was due
to being closer to the basket, which requires more force in an upward direction
in order to get the proper arc. These velocities change relative to the
distance between the shooter and the basket; if the shooter is close to the
basket s/he needs a higher y velocity for the arc, as s/he moves further out
the higher velocity for the x is required, this continues on until a certain
point (depending on the strength of the shooter) where both the x and y will
have to be nearly equal to balance the distance from the hoop and the arc required.
From the tracker we were able to find
the velocity in the xy coordinate planes which was 8.66m/s. This shows that the
ball to hit a 10ft high target at 10ft required the ball to be propelled at a
velocity of 19.3 miles per hour.
The next variable we found necessary
in order to determine the other required variable such as momentum, impulse force,
and work was acceleration. We then went about finding the acceleration of the
ball by using the tracker and the high speed once again. This gave us the
acceleration of the x as 0m/s2, in the y as 10.353m/s2, and in the xy planes
10.353 m/s2. Based on residual memories
of learning physics in 1^{st} semester the acceleration of the ball in
the x plane is 0.0m/s2 because, after the ball leaves the shooters hands the
only force acting on it are gravity and air resistance; due to the latter’s
negligibility and no other force pushing or pulling the ball to give it a
positive or negative acceleration, the acceleration in they is0.0m/s2. In the y plane the acceleration was found to
be 10.353m/s2. This is the velocity in the y plane because as the ball travels
gravity, the only sufficient force acting on the ball, pulls down at a rate of
9.8m/s2 which is added to the decreasing arc of the ball of 2.53 m/s2, both of
which have negative connotations due to their decreasing height. This 10.353
m/s2 is also the total acceleration in the xy plane due to having no x
acceleration and the only acceleration was in the y direction this is then the
xy acceleration as well. Along with the acceleration of the ball we used a wii
mote as an accelerometer to find the acceleration of the shooter predominate
arm. The data received from the wii
mote, was in excess of 55 pages for 2 seconds of shooting, so as a condensed
version one time will be used for each general motion, .as best as can be
determined. As the shooter begins to propel the ball the x,y,and xy
accelerations are respectively ,0.129630,0.000000,1.086207m/s2;
after the ball has left the
shooters hand, which continues upwards, the accelerations are respectively ,0.037037,0.125000,1.034483
m/s2; once the shooters arm stops, the accelerations are respectively, 0.000000,0.000000,0.000000m/s2;
as the arm returns, the accelerations are respectively
,0.185185,0.160714,0.965517m/s2; and as the shooter brings his arms back
to rest the accelerations are respectively 0.018519,0.035714,1.000000m/s2.
These accelerations are all in relation to the movement of the arm, in this
case up and forward are positive while backward and down are negative.
Force: Force is defined as the
push or pull that can cause an object with mass to change velocity. The
equation of force is F=MA , or Force equals Mass x Acceleration. The mass of
the ball, obviously staying constant, is 0.6236895 kg. The acceleration that
was found in Steve doing a jump shot was 10.353 m/s^{2} in the y direction and 0 in the x
direction. Therefore, using the force equation, the overall force of the ball
is 6.457 Newtons. What this means is that about 6 and a half Newtons of force
were acting on the ball was in the air. The reason why there was no force in
the x direction is because the only factor affecting the ball is gravity which
pushes the ball downwards and in no other direction. Therefore the overall
force of the ball is the same as the y force acting on the ball.
Impulse/Momentum: Momentum is
defined as the measure of the motion of an object. The equation for momentum is
mass x velocity. Impulse is defined as the change in momentum, or the force
multiplied by the amount of time the object acts over. The general equation for
impulse is the change in velocity x mass. We found the momentum of the
basketball by multiplying the mass (0.624) and the velocities (4.026 in the x,
5.905 in the y, and 8.662 overall) to get a momentum [in (kg)(m)/s] of 2.508 in
the x, 2.679 in the y and 5.396 overall. The impulse which is the change in
momentum over the time of the ball was found at .0997 in the x, .255 in the y
and .998 overall. What this means is that the change in the momentum was a
result of the change in force due to gravity. Because of the small amount of
time that the shot was recorded, the number did not change very much.
Work/conservation of energy: Work
is defined as the amount of energy transferred by a force acting through a
distance. The common equation for work is W=Fd, or Work equals force times
distance. Based on the data we collected, the Work in the y direction is 3.668
Nm and overall is 5.595 Nm. What this means as that in order for the ball to
be able to move, it required the 5.595 Nm of work to get from the point of
starting to the point that it ended. The law of conservation of energy states
that energy cannot be created or destroyed. The way that these two concepts
relate is that the amount of work that was on the ball came from the work that
Steve applied to the ball when he threw it. The limit of the work applied to
the force can dictate how long the ball would be able to stay in the air. This
is why it is extremely hard to be able to throw a ball even across a basketball
court.
With
only the desire to find the simple physics variable we’ve learned in class
along with a laptop,
lab pro, wii mote as an accelerometer, and a force plate, scale pole, and the
high speed camera, we were able to determine the raw physics necessary for the
common athletic action of a jump shot. .
Building upon the knowledge of the concepts of velocity, acceleration,
mass, force, momentum, impulse, work, and the conservation of energy we learned
in class we were able to devise the plan of measuring those unknowns required
to properly analyze the physics of a jump shot.
For clarifications of what we were thinking while researching the
physics of a jump shot, see the parallel documentary video of the physics of a jump
shot.