In a shotput event an athlete throws the shotput of mass 20 kg with an initial speed of `2 m s^(-1)` at `45^@` from height 3 m above ground. Assuming air resistance to be negligible and acceleration due to gravity to be `10 m s^(-2)`, the kinetic energy of the shotput when it just reaches the ground will be

Options

A. 2.5 J
B. 5.0 J
C. 52.5 J
D. 155.0 J

Explanation

To solve the problem, we will use the principle of conservation of mechanical energy. The total mechanical energy at the initial position (when the shotput is thrown) will be equal to the total mechanical energy just before it hits the ground.

### Step 1: Calculate Initial Kinetic Energy (KE_initial)
The initial kinetic energy can be calculated using the formula:
\[ KE_{\text{initial}} = \frac{1}{2} m u^2 \]
where:
- \( m = 20 \, \text{kg} \) (mass of the shotput)
- \( u = 2 \, \text{m/s} \) (initial speed)

Substituting the values:
\[ KE_{\text{initial}} = \frac{1}{2} \times 20 \times (2)^2 = \frac{1}{2} \times 20 \times 4 = 40 \, \text{J} \]

### Step 2: Calculate Initial Potential Energy (PE_initial)
The initial potential energy can be calculated using the formula:
\[ PE_{\text{initial}} = mgh \]
where:
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h = 3 \, \text{m} \) (height above ground)

Substituting the values:
\[ PE_{\text{initial}} = 20 \times 10 \times 3 = 600 \, \text{J} \]

### Step 3: Calculate Total Initial Energy (E_initial)
The total initial energy is the sum of the initial kinetic energy and the initial potential energy:
\[ E_{\text{initial}} = KE_{\text{initial}} + PE_{\text{initial}} = 40 \, \text{J} + 600 \, \text{J} = 640 \, \text{J} \]

### Step 4: Calculate Final Kinetic Energy (KE_final)
When the shotput reaches the ground, its potential energy is zero (since height is zero). Therefore, all the initial energy will be converted into kinetic energy:
\[ E_{\text{final}} = KE_{\text{final}} + PE_{\text{final}} \]
Since \( PE_{\text{final}} = 0 \):
\[ E_{\text{final}} = KE_{\text{final}} \]
Thus:
\[ KE_{\text{final}} = E_{\text{initial}} = 640 \, \text{J} \]

### Final Answer
The kinetic energy of the shotput when it just reaches the ground will be:
\[ \boxed{640 \, \text{J}} \]