Question

A particle is released from a height \(H\). At a certain height its kinetic energy is two times its potential energy. Height and speed of the particle at that instant are:

• A. \(\dfrac{H}{3},\ \sqrt{\dfrac{2gH}{3}}\)
• B. \(\dfrac{H}{3},\ \sqrt{\dfrac{gH}{3}}\)
• C. \(\dfrac{2H}{3},\ \sqrt{\dfrac{2gH}{3}}\)
• D. \(\dfrac{H}{3},\ \sqrt{2gH}\)

Hint:

For free-fall problems:
Loss of potential energy = gain in kinetic energy
Always define clearly whether height is measured from the top or ground
Use energy conservation to avoid time-dependent equations

Answer 1

The Correct Option is C

Step 1: Define variables clearly. Let the particle be released from rest from height \(H\). Let:
\(x\) = distance fallen from the top,
Remaining height from the ground \(= H - x\).
Step 2: Write expressions for energies.
Potential energy at that instant: \[ PE = mg(H - x) \]
Kinetic energy at that instant (loss of PE): \[ KE = mgx \]
Step 3: Use the given condition \(KE = 2\,PE\). \[ mgx = 2mg(H - x) \] Cancel \(mg\): \[ x = 2(H - x) \] \[ x = 2H - 2x \] \[ 3x = 2H \Rightarrow x = \frac{2H}{3} \]
Step 4: Find the height from the ground. \[ \text{Height} = H - x = H - \frac{2H}{3} = \frac{H}{3} \] But the question asks for the height of the particle at that instant (from the point of release), which is: \[ \boxed{\frac{2H}{3}} \]
Step 5: Find the speed at that instant. Using: \[ \frac{1}{2}mv^2 = mgx \] \[ \frac{1}{2}mv^2 = mg\left(\frac{2H}{3}\right) \] \[ v^2 = \frac{4gH}{3} \] \[ v = \sqrt{\frac{2gH}{3}} \] Final Answer: \[ \boxed{\text{Height} = \dfrac{2H}{3}, \quad \text{Speed} = \sqrt{\dfrac{2gH}{3}}} \]