A 2.5 kg jackfruit is plucked from a height of 4 m. Calculate its gravitational potential energy, kinetic energy just before hitting the ground, and velocity at impact, assuming negligible air resista

Mass, $m = 2500 \, \text{g} = 2.5 \, \text{kg}$
Height, $h = 4 \, \text{m}$
Acceleration due to gravity, $g = 9.8 \, \text{m/s}^2$

The gravitational potential energy is $\boxed{\text{98 J}}$.

The assumption is that air resistance is negligible, and thus, mechanical energy is conserved during the fall. This means all the initial potential energy is converted into kinetic energy just before hitting the ground.

The kinetic energy of the fruit as it hits the ground is $\boxed{\text{98 J}}$.

The velocity with which it hits the ground is approximately $\boxed{\text{8.85 m/s}}$.

c) The jackfruit comes to rest after hitting the ground and does not bounce back to its original height because the collision with the ground is inelastic. During an inelastic collision, some of the kinetic energy is converted into other forms of energy, such as heat, sound, and energy used to deform the fruit and the ground. This energy is dissipated into the surroundings, so it is not available to propel the fruit back up.

The law shown by this behavior is the Law of Conservation of Energy. While mechanical energy (potential + kinetic) is not conserved due to the energy conversion, the total energy of the system (including heat, sound, and deformation energy) remains conserved.