A mass of 2kg is dropped from a height of 3m.Neglecting air resistance,KE of the stone just b4 it hits the ground is ?

KE = 0.5M*V^2.

V^2 = Vo^2 + 2g*h = 0 + 19.6*3 = 58.8.

and of course, the final KE is equal to the initial PE...

initial PE=mgh=2*9.8*3=58.8Joules

Yes, I agree.

To determine the kinetic energy (KE) of the stone just before it hits the ground, we can use the principle of conservation of energy. The potential energy (PE) stored in an object at a certain height is converted into kinetic energy as it falls.

The formula for potential energy is given by:

PE = m * g * h

where
m is the mass of the object (2 kg),
g is the acceleration due to gravity (9.8 m/s²),
and h is the height (3 m).

By substituting the values into the equation, we can calculate the potential energy:

PE = 2 kg * 9.8 m/s² * 3 m
= 58.8 Joules

Since the potential energy is converted entirely into kinetic energy when the object reaches the ground, the kinetic energy just before hitting the ground is equal to the potential energy.

Therefore, the kinetic energy of the stone just before it hits the ground is 58.8 Joules.