Ball A is thrown straight down off the top of a tall building with an initial velocity of 20m/s. Ball b is thrown straight up from the top of the same building with initial velocity of 20 m/s. Which ball, if either, experiences the greater acceleration after leaving the throwers hand?
They both experience the same gravitational force, m g
their acceleration is therefore F/m = g
Period, no other force, no other acceleration.
To determine which ball experiences the greater acceleration after leaving the thrower's hand, we need to consider the forces acting on both balls.
When Ball A is thrown straight down, the force of gravity, acting in the downward direction, causes it to accelerate in the same direction as the initial velocity. Since the velocity and acceleration have the same direction, the magnitude of the acceleration for Ball A remains constant, which is equal to the acceleration due to gravity (9.8 m/s²). Hence, the acceleration of Ball A is 9.8 m/s².
When Ball B is thrown straight up, the force of gravity still acts downward, but in this case, it opposes the initial velocity. Initially, the velocity is upward, and as the force of gravity slows Ball B down, it eventually stops and starts moving downward. As a result, the velocity and acceleration have opposite directions (velocity upwards, acceleration downwards). The magnitude of the acceleration for Ball B remains constant as well, equal to the acceleration due to gravity (9.8 m/s²).
Therefore, both Ball A and Ball B experience the same magnitude of acceleration (9.8 m/s²) after leaving the thrower's hand because they are both subject to the gravitational force, regardless of the direction of their initial velocities.