To solve this problem, we can use the principle of conservation of energy, which tells us the energy in a closed system is constant. When the object is released from rest, it has potential energy that converts to kinetic energy as it falls. When it hits the ground, all the potential energy has been converted to kinetic energy. Assuming there is no air resistance, the potential energy (PE) at the point of release equals the kinetic energy (KE) just before impact.
The formula for gravitational potential energy is:
\[ PE = mgh \]
where:
- \( m \) is the mass of the object,
- \( g \) is the acceleration due to gravity (approximately \( 9.8 \) meters per second squared on Earth), and
- \( h \) is the height from which the object falls.
The formula for kinetic energy is:
\[ KE = \frac{1}{2}mv^2 \]
where:
- \( m \) is the mass of the object,
- \( v \) is the velocity of the object just before impact.
At the point just before impact, the potential energy at the release point has been converted to kinetic energy:
\[ PE_{release} = KE_{impact} \]
Substituting the formulas, we get:
\[ mgh = \frac{1}{2}mv^2 \]
From the problem, we know that \( KE_{impact} = 1960 \) joules, and we are trying to find \( h \).
By conservation of energy and because the mass of the object is the same on both sides of the equation, it cancels out:
\[ gh = \frac{1}{2}v^2 \]
Since we don't have the mass or the velocity of the object, we don't actually need them because we can use the given kinetic energy to find \( h \) directly:
\[ h = \frac{2 \cdot KE}{g} \]
Substituting the given kinetic energy and the acceleration due to gravity, we have:
\[ h = \frac{2 \cdot 1960 \, \text{J}}{9.8 \, \text{m/s}^2} \]
\[ h = \frac{3920 \, \text{J}}{9.8 \, \text{m/s}^2} \]
\[ h \approx 400 \, \text{m} \]
So, the object was released from approximately 400 meters above the ground.