The correct response is: She is right! The higher the drop, the greater kinetic the ball will have upon hitting the ground.
When an object is dropped from a height, it gains kinetic energy as it falls due to the conversion of potential energy to kinetic energy. The potential energy at the height can be calculated using the formula \( PE = mgh \), where \( m \) is the mass, \( g \) is the acceleration due to gravity (approximately 9.81 m/s²), and \( h \) is the height (3.0 meters in this case).
As the ball falls, this potential energy converts into kinetic energy, and by the time it hits the ground, its kinetic energy will be equal to the potential energy it had at the drop height, assuming no energy is lost to air resistance.
To find the kinetic energy when dropped from 3.0 meters, you would calculate:
\[ KE = mgh \]
For a mass \( m \) and \( g = 9.81 , \text{m/s}^2 \), the potential and thus the kinetic energy when reaching the ground will indeed likely be higher than 96 joules for reasonable values of mass. Thus, the student's claim is incorrect if they believe it will be less than that.