A particle is released from a height S. At certain height its kinetic energy is three times its potential energy. The height and speed of the particle at that instant are respectively

Options

A. `S/4, (3gS)/2`
B. `S/4, (sqrt(3gS))/2`
C. `S/2, (sqrt(3gS))/g`
D. `S/4,sqrt((3gS)/2)`

Answer

To solve the problem step by step, we need to analyze the situation where a particle is released from a height \( S \) and at a certain height \( h \), its kinetic energy (KE) is three times its potential energy (PE).

### Step 1: Define the energies at height \( h \)
1. The potential energy at height \( h \) is given by:
\[ PE = mgh \]
2. The kinetic energy at height \( h \) is given by:
\[ KE = \frac{1}{2} mv^2 \]

### Step 2: Relate kinetic and potential energy
From the problem, we know that:
\[ KE = 3 \times PE \]
Substituting the expressions for KE and PE, we get:
\[ \frac{1}{2} mv^2 = 3(mgh) \]

### Step 3: Simplify the equation
Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{1}{2} v^2 = 3gh \]
Multiplying both sides by 2 gives:
\[ v^2 = 6gh \]

### Step 4: Determine the height fallen
The particle is released from height \( S \) and has fallen a distance of \( S - h \). The potential energy lost is:
\[ PE_{\text{lost}} = mg(S - h) \]
This lost potential energy is converted into kinetic energy:
\[ KE = mg(S - h) \]

### Step 5: Set up the equation for potential energy lost
Since we already established that \( KE = 3(mgh) \), we can equate the two expressions for kinetic energy:
\[ mg(S - h) = 3(mgh) \]
Cancel \( mg \) from both sides:
\[ S - h = 3h \]
Rearranging gives:
\[ S = 4h \]
Thus, we can find \( h \):
\[ h = \frac{S}{4} \]

### Step 6: Calculate the speed \( v \)
Now substitute \( h = \frac{S}{4} \) back into the equation for \( v^2 \):
\[ v^2 = 6g\left(\frac{S}{4}\right) = \frac{6gS}{4} = \frac{3gS}{2} \]
Taking the square root gives:
\[ v = \sqrt{\frac{3gS}{2}} \]

### Final Answers
- The height \( h \) at which the kinetic energy is three times the potential energy is:
\[ h = \frac{S}{4} \]
- The speed \( v \) of the particle at that height is:
\[ v = \sqrt{\frac{3gS}{2}} \]