A stone is thrown upwards with a velocity `50 ms^-1`. Another stone is simultaneously thrown downwards from the same location with a velocity `50 ms^-1`. When the first stone is at the highest point, the velocity of the second stone is `("Take" g = 10^-2)` :

Options

A. Zero
B. `50 ms^-1`
C. `100 ms^-1`
D. `150 ms^-1`

Explanation

To solve the problem, we need to analyze the motion of both stones separately.

### Step-by-Step Solution:

1. **Identify the Motion of the First Stone**:
- The first stone is thrown upwards with an initial velocity \( u_1 = 50 \, \text{m/s} \).
- When it reaches its highest point, its final velocity \( v_1 = 0 \, \text{m/s} \).
- The acceleration due to gravity \( g = 10 \, \text{m/s}^2 \) acts downwards, so we take it as negative in our calculations.

2. **Use the Equation of Motion for the First Stone**:
- We can use the equation:
\[ v = u + at \]
- Here, \( v = 0 \), \( u = 50 \, \text{m/s} \), and \( a = -g = -10 \, \text{m/s}^2 \).
- Plugging in the values:
\[ 0 = 50 - 10t \]
- Rearranging gives:
\[ 10t = 50 \implies t = 5 \, \text{s} \]
- Thus, the time taken for the first stone to reach the highest point is \( 5 \, \text{s} \).

3. **Identify the Motion of the Second Stone**:
- The second stone is thrown downwards with an initial velocity \( u_2 = -50 \, \text{m/s} \) (negative because it is in the downward direction).
- We need to find its velocity when the first stone is at its highest point, which is after \( 5 \, \text{s} \).

4. **Use the Equation of Motion for the Second Stone**:
- We again use the equation of motion:
\[ v = u + at \]
- Here, \( u = -50 \, \text{m/s} \), \( a = g = 10 \, \text{m/s}^2 \), and \( t = 5 \, \text{s} \).
- Plugging in the values:
\[ v = -50 + 10 \times 5 \]
- Simplifying gives:
\[ v = -50 + 50 = 0 \, \text{m/s} \]
- However, we need to consider the total distance fallen by the second stone during the 5 seconds.

5. **Calculate the Distance Fallen by the Second Stone**:
- The distance fallen can be calculated using:
\[ s = ut + \frac{1}{2} a t^2 \]
- Plugging in the values:
\[ s = (-50)(5) + \frac{1}{2}(10)(5^2) \]
- This simplifies to:
\[ s = -250 + \frac{1}{2}(10)(25) = -250 + 125 = -125 \, \text{m} \]
- The second stone has fallen \( 125 \, \text{m} \) downwards.

6. **Calculate the Final Velocity of the Second Stone**:
- Now, we can find the final velocity using:
\[ v = u + at \]
- We already have \( u = -50 \, \text{m/s} \), \( a = 10 \, \text{m/s}^2 \), and \( t = 5 \, \text{s} \):
\[ v = -50 + 10 \times 5 = -50 + 50 = 0 \, \text{m/s} \]
- However, we need to calculate the velocity at the time it has fallen \( 125 \, \text{m} \):
\[ v^2 = u^2 + 2as \]
- Plugging in the values:
\[ v^2 = (-50)^2 + 2(10)(125) \]
- This simplifies to:
\[ v^2 = 2500 + 2500 = 5000 \]
- Therefore, \( v = \sqrt{5000} = 70.71 \, \text{m/s} \) downwards.

### Final Answer:
The velocity of the second stone when the first stone is at the highest point is approximately \( 70.71 \, \text{m/s} \) downwards.