A ball is thrown vertically upwards with a speed of `10 m//s` from the top of a tower 200 m height and another is thrown vertically downwards with the same speed simultaneously. The time difference between them on reaching the ground is `(g=10 m//s^(2))`

Options

A. 12 s
B. 6 s
C. 2 s
D. 1 s

Explanation

To solve the problem, we need to analyze the motion of both balls: one thrown upwards from the top of a 200 m tower and the other thrown downwards from the same height. We will calculate the time taken by each ball to reach the ground and then find the time difference between their arrivals.

### Step-by-step Solution:

1. **Identify the Given Data:**
- Height of the tower (h) = 200 m
- Initial speed of both balls (u) = 10 m/s
- Acceleration due to gravity (g) = 10 m/s²

2. **Calculate the Time Taken by the Ball Thrown Upwards:**
- When the ball is thrown upwards, it will first rise to a maximum height before falling back down.
- The time taken to reach the maximum height (where final velocity = 0) can be calculated using the equation:
\[ v = u + at \]
Here, \(v = 0\) (at maximum height), \(u = 10 \, \text{m/s}\), and \(a = -g = -10 \, \text{m/s}^2\).
\[ 0 = 10 - 10t \implies t = 1 \, \text{s} \]
- The ball will take 1 second to reach the maximum height.

3. **Calculate the Maximum Height Reached:**
- The maximum height (H) reached by the ball can be calculated using:
\[ H = ut + \frac{1}{2} a t^2 \]
\[ H = 10(1) + \frac{1}{2}(-10)(1^2) = 10 - 5 = 5 \, \text{m} \]
- Therefore, the total height from the ground when the ball is at its maximum height is:
\[ \text{Total height} = 200 \, \text{m} + 5 \, \text{m} = 205 \, \text{m} \]

4. **Calculate the Time Taken to Fall from Maximum Height to Ground:**
- Now, we need to calculate the time taken to fall from 205 m to the ground using the equation:
\[ h = \frac{1}{2} g t^2 \]
\[ 205 = \frac{1}{2}(10)t^2 \implies 205 = 5t^2 \implies t^2 = 41 \implies t = \sqrt{41} \approx 6.4 \, \text{s} \]

5. **Total Time for the Upward Thrown Ball:**
- Total time for the ball thrown upwards:
\[ T_{up} = 1 \, \text{s} + 6.4 \, \text{s} = 7.4 \, \text{s} \]

6. **Calculate the Time Taken by the Ball Thrown Downwards:**
- The ball thrown downwards has an initial velocity of 10 m/s and falls from 200 m. We can use the same equation:
\[ h = ut + \frac{1}{2} g t^2 \]
\[ 200 = 10t + \frac{1}{2}(10)t^2 \implies 200 = 10t + 5t^2 \]
Rearranging gives:
\[ 5t^2 + 10t - 200 = 0 \]
Dividing the entire equation by 5:
\[ t^2 + 2t - 40 = 0 \]
Using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ t = \frac{-2 \pm \sqrt{(2)^2 - 4(1)(-40)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 160}}{2} = \frac{-2 \pm 12.65}{2} \]
Taking the positive root:
\[ t \approx 5.3 \, \text{s} \]

7. **Calculate the Time Difference:**
- The time difference between the two balls reaching the ground is:
\[ \Delta t = T_{up} - T_{down} = 7.4 \, \text{s} - 5.3 \, \text{s} = 2.1 \, \text{s} \]

### Final Answer:
The time difference between the two balls reaching the ground is approximately **2.1 seconds**.