She is incorrect because the higher the ball is dropped from, the greater kinetic energy it will have when it hits the ground.
When an object is dropped from a height, its potential energy at the start is converted into kinetic energy as it falls. The potential energy (PE) can be calculated using the formula:
\[ PE = mgh \]
where
- \( m \) is the mass of the object (in kilograms),
- \( g \) is the acceleration due to gravity (approximately \( 9.81 , \text{m/s}^2 \)),
- \( h \) is the height from which it is dropped (in meters).
In this case, we can determine the maximum potential energy just before hitting the ground by assuming the mass to be such that it results in a total potential energy greater than 96 joules.
For example, if we consider a 1 kg ball:
\[ PE = 1 , \text{kg} \times 9.81 , \text{m/s}^2 \times 3.0 , \text{m} \] \[ PE = 29.43 , \text{joules} \]
Even for a 3 kg ball:
\[ PE = 3 , \text{kg} \times 9.81 , \text{m/s}^2 \times 3.0 , \text{m} \] \[ PE = 88.29 , \text{joules} \]
It is important to note that to exceed 96 joules, the mass would need to be approximately 3.3 kg, as calculated:
\[ 96 , \text{joules} = m \times 9.81 \times 3 \] \[ m = \frac{96}{29.43} \approx 3.26 , \text{kg} \]
Thus, if the ball’s mass is less than about 3.3 kg, it will have less than 96 joules of kinetic energy upon hitting the ground. Therefore, while the potential energy can vary based on the mass, the student’s assertion that a ball dropped from 3.0 meters has less than 96 joules of kinetic energy is not universally correct, as it depends on the mass of the ball.