A particle is released from height `H`. At certain height from the ground its kinetic energy is twice its gravitational potential energy. Find the height and speed of particle at that height.

Options

A. `(H)/(3),sqrt((2gH)/(3))`
B. `(H)/(3),2sqrt((gH)/(3))`
C. `(2H)/(3),sqrt((2gH)/(3))`
D. `(H)/(3),sqrt(2gH)`

Explanation

To solve the problem step by step, we will use the principles of conservation of energy and the relationships between kinetic energy (KE) and gravitational potential energy (PE).

### Step 1: Understand the problem
A particle is released from a height \( H \). At a certain height \( x \) above the ground, its kinetic energy is twice its gravitational potential energy. We need to find the height \( x \) and the speed \( v \) of the particle at that height.

### Step 2: Write the expressions for kinetic and potential energy
The gravitational potential energy (PE) at height \( x \) is given by:
\[ PE = m g x \]
The kinetic energy (KE) at that height is given by:
\[ KE = 2 \times PE = 2 m g x \]

### Step 3: Apply conservation of energy
The total mechanical energy at the height \( H \) when the particle is released is equal to the sum of kinetic and potential energy at height \( x \):
\[ m g H = KE + PE \]
Substituting the expressions for KE and PE:
\[ m g H = 2 m g x + m g x \]
This simplifies to:
\[ m g H = 3 m g x \]

### Step 4: Cancel the common terms
We can cancel \( m g \) from both sides (assuming \( m \neq 0 \) and \( g \neq 0 \)):
\[ H = 3 x \]

### Step 5: Solve for height \( x \)
Rearranging the equation gives:
\[ x = \frac{H}{3} \]

### Step 6: Find the speed \( v \) at height \( x \)
We know that the kinetic energy can also be expressed in terms of speed:
\[ KE = \frac{1}{2} m v^2 \]
Setting this equal to the expression for kinetic energy we derived earlier:
\[ \frac{1}{2} m v^2 = 2 m g x \]
Again, cancel \( m \) from both sides:
\[ \frac{1}{2} v^2 = 2 g x \]
Multiplying both sides by 2 gives:
\[ v^2 = 4 g x \]

### Step 7: Substitute \( x \) into the equation
Now substitute \( x = \frac{H}{3} \) into the equation for \( v^2 \):
\[ v^2 = 4 g \left(\frac{H}{3}\right) \]
This simplifies to:
\[ v^2 = \frac{4 g H}{3} \]

### Step 8: Solve for speed \( v \)
Taking the square root of both sides gives:
\[ v = \sqrt{\frac{4 g H}{3}} = 2 \sqrt{\frac{g H}{3}} \]

### Final Results
Thus, the height \( x \) and the speed \( v \) of the particle at that height are:
- Height \( x = \frac{H}{3} \)
- Speed \( v = 2 \sqrt{\frac{g H}{3}} \)