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 will analyze the situation of a particle released from a height \( S \) and find the height and speed of the particle when its kinetic energy is three times its potential energy.

### Step 1: Understand the Energy Conservation
When the particle is released from height \( S \), its total mechanical energy is conserved. The total mechanical energy at the height \( S \) is given by:
\[ E_{total} = mgh = mgS \]
where \( m \) is the mass of the particle and \( g \) is the acceleration due to gravity.

**Hint:** Remember that total mechanical energy remains constant in the absence of air resistance.

### Step 2: Define Kinetic and Potential Energy
At a certain height \( h \) (where \( h < S \)), the potential energy \( PE \) and kinetic energy \( KE \) of the particle can be expressed as:
\[ PE = mgh \]
\[ KE = \frac{1}{2}mv^2 \]
According to the problem, the kinetic energy is three times the potential energy:
\[ KE = 3PE \]

**Hint:** Use the relationship between kinetic and potential energy to set up your equations.

### Step 3: Set Up the Equation
Substituting the expressions for \( KE \) and \( PE \):
\[ \frac{1}{2}mv^2 = 3(mgh) \]
This simplifies to:
\[ \frac{1}{2}mv^2 = 3mg( S - h) \]
Here, \( S - h \) is the height fallen from the original height \( S \).

**Hint:** Remember that the potential energy decreases as the particle falls, while kinetic energy increases.

### Step 4: Express Total Energy
From conservation of energy, the total energy at height \( h \) can also be expressed as:
\[ E_{total} = KE + PE = \frac{1}{2}mv^2 + mgh \]
Setting this equal to the total energy at height \( S \):
\[ mgS = \frac{1}{2}mv^2 + mgh \]

**Hint:** This equation helps relate the height fallen and the remaining height.

### Step 5: Substitute and Solve for Height
From the earlier equation \( \frac{1}{2}mv^2 = 3mg(S - h) \), we can express \( v^2 \):
\[ v^2 = 6g(S - h) \]
Now substituting \( v^2 \) into the total energy equation:
\[ mgS = \frac{1}{2}m(6g(S - h)) + mgh \]
This simplifies to:
\[ mgS = 3mg(S - h) + mgh \]
Dividing through by \( mg \):
\[ S = 3(S - h) + h \]
Expanding and rearranging gives:
\[ S = 3S - 3h + h \]
\[ S = 3S - 2h \]
\[ 2h = 2S \implies h = \frac{S}{4} \]

**Hint:** Keep track of your algebraic manipulations to avoid mistakes.

### Step 6: Calculate Speed
Now, substituting \( h = \frac{S}{4} \) back into the expression for \( v^2 \):
\[ v^2 = 6g\left(S - \frac{S}{4}\right) = 6g\left(\frac{3S}{4}\right) = \frac{18gS}{4} = \frac{9gS}{2} \]
Thus, the speed \( v \) is:
\[ v = \sqrt{\frac{9gS}{2}} = \frac{3\sqrt{gS}}{\sqrt{2}} \]

**Hint:** Ensure you understand how to derive speed from kinetic energy.

### Final Answer
The height \( h \) is \( \frac{S}{4} \) and the speed \( v \) is \( \frac{3\sqrt{gS}}{\sqrt{2}} \).