A Superball is dropped from a height of 2.4 and rebounds off the floor to a height of 1.9 .
Part A
If the ball is in contact with the floor for 0.70 , determine the direction of the ball’s average acceleration due to the floor.
If the ball is in contact with the floor for 0.70 , determine the direction of the ball’s average acceleration due to the floor.
downward
upward
Part B
Determine the magnitude of the ball’s average acceleration due to the floor.
Express your answer using two significant figures.
To determine the direction of the ball's average acceleration due to the floor, we need to consider the change in velocity during the contact time.
Given that the ball was dropped from a height of 2.4 m and rebounded to a height of 1.9 m, we can conclude that the ball's final velocity is less than its initial velocity. This means that the ball is slowing down during its contact with the floor.
Since the ball is slowing down, its average acceleration due to the floor must be in the opposite direction of its initial velocity. Considering that the initial direction of the ball's motion is downward (because it was dropped), we can conclude that the direction of the ball's average acceleration due to the floor is upward.
Answer: The direction of the ball's average acceleration due to the floor is upward.
To determine the magnitude of the ball's average acceleration due to the floor, we can use the following kinematic equation:
velocity^2 = initial_velocity^2 + 2 * acceleration * displacement
Since we are only interested in the magnitude of acceleration, we can rearrange the equation as:
acceleration = (velocity^2 - initial_velocity^2) / (2 * displacement)
Given that the ball is in contact with the floor for 0.70 s, we can calculate the initial velocity (vi) and the final velocity (vf) during this time:
vi = 0 m/s (since the ball is dropped from rest)
vf = (final_position - initial_position) / contact_time
= (1.9 m - 2.4 m) / 0.70 s
Substituting the values into the equation, we can calculate the magnitude of the ball's average acceleration due to the floor.