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. 2.5J
B. 5J
C. 525J
D. 640 J

Explanation

To find the kinetic energy of the shotput when it just reaches the ground, we can use the principle of conservation of mechanical energy. The total mechanical energy at the initial point (when the shotput is thrown) will be equal to the total mechanical energy just before it hits the ground.

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- Mass of the shotput, \( m = 20 \, \text{kg} \)
- Initial speed, \( v_0 = 2 \, \text{m/s} \)
- Angle of projection, \( \theta = 45^\circ \)
- Height from which it is thrown, \( h = 3 \, \text{m} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Calculate Initial Kinetic Energy (KE_initial):**
The kinetic energy is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
Substituting the values:
\[ KE_{\text{initial}} = \frac{1}{2} \times 20 \, \text{kg} \times (2 \, \text{m/s})^2 \]
\[ KE_{\text{initial}} = \frac{1}{2} \times 20 \times 4 = 40 \, \text{J} \]

3. **Calculate Initial Potential Energy (PE_initial):**
The potential energy is given by the formula:
\[ PE = mgh \]
Substituting the values:
\[ PE_{\text{initial}} = 20 \, \text{kg} \times 10 \, \text{m/s}^2 \times 3 \, \text{m} \]
\[ PE_{\text{initial}} = 20 \times 10 \times 3 = 600 \, \text{J} \]

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

5. **Final Conditions Just Before Hitting the Ground:**
Just before the shotput hits the ground, its potential energy will be zero (since height = 0). Therefore, all the initial energy will convert into kinetic energy:
\[ KE_{\text{final}} = E_{\text{initial}} = 640 \, \text{J} \]

### Conclusion:
The kinetic energy of the shotput when it just reaches the ground will be \( 640 \, \text{J} \).

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