The Determination of the Ideal Velocity of the Bouncing Ball
A hard rubber ball is dropped from rest. It falls to the concrete floor and bounces back up ALMOST to its initial height. A motion detector is mounted on the ceiling directly above the ball, facing down. So, the positive direction -- the away-from-the-detector direction -- is downward. Draw the position, velocity, and acceleration graphs.
Since moving downward is a positive direction, is the velocity positive when the ball falls, 0 when it hits the ground, and then negative when it moves up? Is the accerlation positive (9.8) as the ball falls, very positive when it hits the ground, and then neg (almost -9.8) when it moves up? Keep in mind that the problem says: the ball bounces back up ALMOST to its initial height. Also, please help me with understanding how the position graph should look like. If there is any way that you could draw me the graphs, that would really be helpful. Thank you!
You're halfway right.
Velocity will be positive as the ball falls, zero on impact, negative on the way up, and zero again at apogee.
Acceleration, however, is constantly 9.8 m/s²...the force of gravity does not change as the ball moves.
For your graphs, acceleration will be a horizontal line at y=9.8
Velocity will be a broken sinusiod (oscillating) that starts at the origin and rises to a maximum. On each impact with the ground, the graph will drop straight down to a minimum, then rise gradually through zero back up to its next maximum which will be less than the previous maximum. This will repeat until the ball is at rest.
Position will be a smooth sinusiod that begins at a maximum, falls to zero as the ball hits the ground, then rises up again to the next maximum which will be less than the previous one. This wave repeats until the ball comes to rest. This graph will never be negative, since the ball never falls below ground level.
Almost everybody, at some point in their lives, has bounced a rubber ball against the wall or floor and observed its motion.
Since moving downward is a positive direction, is the velocity positive when the ball falls, 0 when it hits the ground, and then negative when it moves up? Is the accerlation positive (9.8) as the ball falls, very positive when it hits the ground, and then neg (almost -9.8) when it moves up? Keep in mind that the problem says: the ball bounces back up ALMOST to its initial height. Also, please help me with understanding how the position graph should look like. If there is any way that you could draw me the graphs, that would really be helpful. Thank you!
You're halfway right.
Velocity will be positive as the ball falls, zero on impact, negative on the way up, and zero again at apogee.
Acceleration, however, is constantly 9.8 m/s²...the force of gravity does not change as the ball moves.
For your graphs, acceleration will be a horizontal line at y=9.8
Velocity will be a broken sinusiod (oscillating) that starts at the origin and rises to a maximum. On each impact with the ground, the graph will drop straight down to a minimum, then rise gradually through zero back up to its next maximum which will be less than the previous maximum. This will repeat until the ball is at rest.
Position will be a smooth sinusiod that begins at a maximum, falls to zero as the ball hits the ground, then rises up again to the next maximum which will be less than the previous one. This wave repeats until the ball comes to rest. This graph will never be negative, since the ball never falls below ground level.
Almost everybody, at some point in their lives, has bounced a rubber ball against the wall or floor and observed its motion.