Step 1 : Understanding the Question:
This problem deals with the motion of an object undergoing free fall under the influence of a uniform gravitational field. We are asked to determine the final kinetic energy of a body of mass \( m \) that is dropped from rest at an initial height \( h \), right before it reaches the ground.
Step 2 : Key Formulas and Approach:
The most straightforward approach to solve this problem is to apply the Law of Conservation of Mechanical Energy. This principle states that the total mechanical energy in an isolated system remains constant if only conservative forces (such as gravity) are acting.
The total mechanical energy \( E_{\text{total}} \) at any point is given by:
\[ E_{\text{total}} = KE + PE \]
where the kinetic energy is \( KE = \frac{1}{2}mv^2 \) and the gravitational potential energy is \( PE = mgh \).
Step 3 : Detailed Explanation:
At the release height \( h \), the body is at rest because it is "dropped", meaning its initial velocity is zero. Consequently, its initial kinetic energy \( KE_i \) is equal to \( 0 \).
At this peak height, the gravitational potential energy of the mass is at its maximum, which is expressed as \( PE_i = mgh \).
Summing these individual components gives the total initial mechanical energy of the system: \( E_i = PE_i + KE_i = mgh + 0 = mgh \).
As the body falls freely under gravity, its potential energy decreases as it is converted entirely into kinetic energy.
Just before making contact with the ground, the height of the body is zero, which means the final potential energy \( PE_f \) becomes \( 0 \).
The final mechanical energy is therefore composed entirely of kinetic energy: \( E_f = PE_f + KE_f = 0 + KE_f \).
Because the total mechanical energy of the system must remain constant (\( E_i = E_f \)), we can equate the two states: \( mgh = KE_f \).
Step 4 : Final Answer:
The kinetic energy of the body just before it hits the ground is \( mgh \), which corresponds to option (A).