A ball of mass 50 g is thrown at a speed of 200 m `s^(-1)` due to atmospheric resistances its kinetic energy reduces to 10 % of its initial kinetic energy. Now the speed of ball is
(Consider zero gravity)

Options

A. `2 sqrt10ms^(-1)`
B. `20 sqrt10ms^(-1)`
C. `10 sqrt2ms^(-1)`
D. `10 sqrt20ms^(-1)`

Explanation

To solve the problem, we need to find the final speed of the ball after its kinetic energy has reduced to 10% of its initial kinetic energy due to atmospheric resistance.

### Step-by-step Solution:

1. **Identify the initial parameters:**
- Mass of the ball, \( m = 50 \, \text{g} = 50 \times 10^{-3} \, \text{kg} = 0.05 \, \text{kg} \)
- Initial speed of the ball, \( v_i = 200 \, \text{m/s} \)

2. **Calculate the initial kinetic energy (KE_initial):**
\[ KE_{\text{initial}} = \frac{1}{2} m v_i^2 \]
Substituting the values:
\[ KE_{\text{initial}} = \frac{1}{2} \times 0.05 \times (200)^2 = \frac{1}{2} \times 0.05 \times 40000 = 1000 \, \text{J} \]

3. **Determine the final kinetic energy (KE_final):**
Since the final kinetic energy is 10% of the initial kinetic energy:
\[ KE_{\text{final}} = 0.1 \times KE_{\text{initial}} = 0.1 \times 1000 = 100 \, \text{J} \]

4. **Set up the equation for final kinetic energy:**
\[ KE_{\text{final}} = \frac{1}{2} m v_f^2 \]
Where \( v_f \) is the final speed of the ball. Substituting the values:
\[ 100 = \frac{1}{2} \times 0.05 \times v_f^2 \]

5. **Solve for \( v_f^2 \):**
Rearranging the equation:
\[ v_f^2 = \frac{100 \times 2}{0.05} = \frac{200}{0.05} = 4000 \]

6. **Calculate \( v_f \):**
Taking the square root:
\[ v_f = \sqrt{4000} = 20\sqrt{10} \, \text{m/s} \]

### Final Answer:
The speed of the ball after the kinetic energy has reduced is \( v_f = 20\sqrt{10} \, \text{m/s} \).

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