A stone of mass 2 kg is projected upwards with KE of 98 J. The height at which the KE of the body becomes half its original value, is given by (take, `g=9.8ms^(-2)`)

Options

A. 5 m
B. 2.5 m
C. 1.5 m
D. 0.5 m

Explanation

To solve the problem step by step, we will use the principles of energy conservation. The kinetic energy of the stone will be converted into potential energy as it rises.

### Step 1: Identify the given values
- Mass of the stone, \( m = 2 \, \text{kg} \)
- Initial kinetic energy, \( KE = 98 \, \text{J} \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

### Step 2: Calculate half of the initial kinetic energy
We need to find the height at which the kinetic energy becomes half of its original value:
\[ KE_{\text{half}} = \frac{KE}{2} = \frac{98 \, \text{J}}{2} = 49 \, \text{J} \]

### Step 3: Set up the equation for potential energy
At the height where the kinetic energy is half, the potential energy gained by the stone will be equal to the kinetic energy lost:
\[ PE = mgh \]
Where \( PE \) is the potential energy, \( m \) is the mass, \( g \) is the acceleration due to gravity, and \( h \) is the height.

### Step 4: Set the potential energy equal to half the kinetic energy
\[ mgh = KE_{\text{half}} \]
Substituting the known values:
\[ 2 \cdot 9.8 \cdot h = 49 \]

### Step 5: Solve for height \( h \)
Now, we can solve for \( h \):
\[ 19.6h = 49 \]
\[ h = \frac{49}{19.6} \]
Calculating \( h \):
\[ h = 2.5 \, \text{m} \]

### Final Answer
The height at which the kinetic energy of the body becomes half its original value is \( h = 2.5 \, \text{m} \).

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