Question

A body of mass \( m \) is dropped from a height \( h \). What is its kinetic energy just before it hits the ground?

• A. \( mgh \)
• B. \( \frac{1}{2} mgh \)
• C. \( 2mgh \)
• D. \( mgh^2 \)

Hint:

For objects in free fall, the potential energy lost equals the kinetic energy gained.

Answer 1

The Correct Option is A

Concept: According to the Law of Conservation of Mechanical Energy, the total mechanical energy in a closed system remains constant if only conservative forces (like gravity) are acting on the body.

Step 1: Analyze the initial energy of the system. When the body is at height \( h \), its velocity is zero (it is "dropped"). Therefore, its initial kinetic energy ($KE_i$) is 0, and its initial potential energy ($PE_i$) is $mgh$. \[ E_{\text{total}} = PE_i + KE_i = mgh + 0 = mgh \]

Step 2: Analyze the energy just before impact. Just before the body hits the ground, its height is 0, meaning the potential energy ($PE_f$) is 0. All the initial potential energy has been converted into kinetic energy ($KE_f$). \[ E_{\text{total}} = PE_f + KE_f = 0 + KE_f \]

Step 3: Equate energies and solve. Since the total energy must remain constant: \[ mgh = 0 + KE_f \implies KE_f = mgh \]