To determine whether the student is correct, we can calculate the potential energy of the ball at the height of 3.0 meters. When the ball is dropped, its potential energy is converted into kinetic energy as it falls. The formula for gravitational potential energy (PE) is:
\[ PE = m \cdot g \cdot h \]
where:
- \(m\) is the mass of the ball (in kilograms),
- \(g\) is the acceleration due to gravity (approximately \(9.81 , \text{m/s}^2\)),
- \(h\) is the height from which the ball is dropped (in meters).
Given that the height \(h = 3.0 , \text{m}\), we need to calculate the potential energy, which will be equal to the kinetic energy (KE) just before it hits the ground (assuming no air resistance).
Substituting in the values we have:
\[ PE = m \cdot 9.81 , \text{m/s}^2 \cdot 3.0 , \text{m} = 29.43m , \text{J} \]
The kinetic energy when the ball hits the ground will be equal to the potential energy at the height of 3.0 m:
\[ KE = PE = 29.43m , \text{J} \]
Now, to find out whether this energy is less than 96 joules, we can analyze the expression \(29.43m , \text{J}\):
- If \(m = 1 , \text{kg}\), then \(KE = 29.43 , \text{J}\).
- If \(m = 2 , \text{kg}\), then \(KE = 58.86 , \text{J}\).
- If \(m = 3 , \text{kg}\), then \(KE = 88.29 , \text{J}\).
- If \(m = 4 , \text{kg}\), then \(KE = 117.72 , \text{J}\).
From the calculations, we can see that for a mass of 3 kg or less, the kinetic energy is less than 96 joules. The kinetic energy will equal or exceed 96 joules when the mass exceeds approximately 3.26 kg (since \(29.43m \geq 96\) yields \(m \geq \frac{96}{29.43} \approx 3.26 , \text{kg}\)).
Thus, the student is correct, assuming the mass of the ball is 3 kg or less, the kinetic energy when it hits the ground will indeed be less than 96 joules. If the mass of the ball is greater than 3.26 kg, the kinetic energy would exceed 96 joules.