Question

A helicopter drops a 25kg of rice from rest height of 120m from the ground what is the kinetic energy of the bag of rice just as it hits the ground​

Answer 1

Answer: 29,403 Joules

Explanation: The kinetic energy of the rice can be calculated using the formula: Kinetic Energy = (1/2) * m *
`v^(2)` where the velocity of the rice just before it hits the ground can be calculated using the formula: v = sqrt(2gh). Substituting the given values, we get: v =
`√(2*9.8 m/s^2 * 120 m)` = 48.5 m/s. Substituting the value of Kinetic Energy = (1/2) * 25 kg *
`48.5 m/s^(2)` = 29,403 J. Therefore, the kinetic energy of the package just before it hits the ground is also 29,403 Joules.

A moving object's kinetic energy is equal to the effort necessary to accelerate it from rest to that speed or the work it can perform while being brought to rest:Since kinetic energy grows with the square of speed, an object moving at twice the speed has four times the kinetic energy as one moving at the rest.