From the top of a tower of height `200 m`, a ball `A` is projected up with `10 m s^(-1)`. And `2 s` later another ball `B` is projected verticall down with the same speed. Then .

Options

A. Both `A` and `B` will reach the ground simultaneously
B. Ball `A` will hit the ground `2 s` later than B hitting the ground.
C. Both the balls will ground with same velocity.
D. Both the balls will hit the ground with different velocity.

Explanation

To solve the problem, we need to analyze the motion of both balls A and B separately and find the time taken for each to reach the ground.

### Step-by-step Solution:

1. **Understanding the Problem:**
- Ball A is projected upwards from the top of a tower of height 200 m with an initial velocity of \( u_A = 10 \, \text{m/s} \).
- Ball B is projected downwards from the same height 2 seconds after Ball A with the same initial velocity \( u_B = 10 \, \text{m/s} \).

2. **Equation of Motion for Ball A:**
- The displacement \( s \) of Ball A when it reaches the ground is -200 m (downward).
- The acceleration \( a \) due to gravity is \( -10 \, \text{m/s}^2 \) (downward).
- The equation of motion we will use is:
\[ s = ut + \frac{1}{2} a t^2 \]
- Plugging in the values for Ball A:
\[ -200 = 10t_A - \frac{1}{2} \cdot 10 \cdot t_A^2 \]
- Rearranging gives:
\[ -200 = 10t_A - 5t_A^2 \]
\[ 5t_A^2 - 10t_A - 200 = 0 \]
- Dividing the entire equation by 5:
\[ t_A^2 - 2t_A - 40 = 0 \]

3. **Solving the Quadratic Equation for Ball A:**
- Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ t_A = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-40)}}{2 \cdot 1} \]
\[ t_A = \frac{2 \pm \sqrt{4 + 160}}{2} \]
\[ t_A = \frac{2 \pm \sqrt{164}}{2} \]
\[ t_A = \frac{2 \pm 12.81}{2} \]
- Taking the positive root:
\[ t_A = \frac{14.81}{2} \approx 7.4 \, \text{s} \]

4. **Equation of Motion for Ball B:**
- Ball B is projected 2 seconds after Ball A, so its time of flight \( t_B \) will be \( t_A - 2 \).
- The displacement for Ball B is also -200 m:
\[ -200 = -10(t_B) - \frac{1}{2} \cdot 10 \cdot t_B^2 \]
- Rearranging gives:
\[ -200 = -10t_B - 5t_B^2 \]
\[ 5t_B^2 + 10t_B - 200 = 0 \]
- Dividing by 5:
\[ t_B^2 + 2t_B - 40 = 0 \]

5. **Solving the Quadratic Equation for Ball B:**
- Using the quadratic formula:
\[ t_B = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 1 \cdot (-40)}}{2 \cdot 1} \]
\[ t_B = \frac{-2 \pm \sqrt{4 + 160}}{2} \]
\[ t_B = \frac{-2 \pm 12.81}{2} \]
- Taking the positive root:
\[ t_B = \frac{10.81}{2} \approx 5.4 \, \text{s} \]

6. **Comparing the Times:**
- Since Ball B is thrown 2 seconds after Ball A, the total time for Ball A is \( 7.4 \, \text{s} \) and for Ball B is \( 5.4 \, \text{s} \).
- The time difference is exactly 2 seconds, which means both balls reach the ground simultaneously.

### Conclusion:
Both Ball A and Ball B will reach the ground at the same time.