jburns4
A Smashing Success by James Burns
James Burns
Contrived
"A Smashing Success"
This photo depicts a medicine ball (which is a weighted ball used for exercise) being dropped from approximately 1 meter onto the wood surface of a racquetball court. The weight of the ball is rated at 7 kilograms. This photo, along with these values, can be used to represent several properties of physics. First, using basic kinematics, we can use the equation Vf^2 = Vi^2 + 2(a)(d) to find the speed of the ball right before it impacts the floor. Where 'Vf' is the final velocity (unknown), 'Vi' is the initial velocity (0 m/s), 'a' is the acceleration due to gravity (9.8 meters/second^2) and 'd' is the vertical displacement (1 meter). After plugging those values in the equation, we find that right before the ball impacts the floor it is moving 4.4 meters per second.
With this, we can then find how much kinetic energy the ball possessed immediately before striking the ground. The equation for this is Kinetic Energy = 1/2(m)v^2, where 'm' is the ball's mass (7 kilograms) and 'v' is the velocity, which we found to be 4.4 meters per second. Again, by simply plugging in these values, we find that the kinetic energy of the ball is 68 Joules. The Joule is a measure of energy.
Because the ball has all this energy when it contacts the ground, much of it is dissipated by the deformation of the structure of the ball as seen in the picture.
A Smashing Success by James Burns
James Burns
Contrived
"A Smashing Success"
This photo depicts a medicine ball (which is a weighted ball used for exercise) being dropped from approximately 1 meter onto the wood surface of a racquetball court. The weight of the ball is rated at 7 kilograms. This photo, along with these values, can be used to represent several properties of physics. First, using basic kinematics, we can use the equation Vf^2 = Vi^2 + 2(a)(d) to find the speed of the ball right before it impacts the floor. Where 'Vf' is the final velocity (unknown), 'Vi' is the initial velocity (0 m/s), 'a' is the acceleration due to gravity (9.8 meters/second^2) and 'd' is the vertical displacement (1 meter). After plugging those values in the equation, we find that right before the ball impacts the floor it is moving 4.4 meters per second.
With this, we can then find how much kinetic energy the ball possessed immediately before striking the ground. The equation for this is Kinetic Energy = 1/2(m)v^2, where 'm' is the ball's mass (7 kilograms) and 'v' is the velocity, which we found to be 4.4 meters per second. Again, by simply plugging in these values, we find that the kinetic energy of the ball is 68 Joules. The Joule is a measure of energy.
Because the ball has all this energy when it contacts the ground, much of it is dissipated by the deformation of the structure of the ball as seen in the picture.