A ball is thrown vertically down with velocity of `5m//s`. With what velocity should another ball be thrown down after 2 seconds so that it can hit the `1^(st)` ball in next 2 second.

Options

A. `40m//s`
B. `10m//s`
C. `15m//s`
D. `20m//s`

Answer

To solve the problem, we need to find the velocity with which the second ball should be thrown downwards after 2 seconds so that it can hit the first ball in the next 2 seconds.

### Step-by-Step Solution:

1. **Understanding the Motion of the First Ball:**
- The first ball is thrown downwards with an initial velocity \( u_1 = 5 \, \text{m/s} \).
- It is in motion for a total time of \( t_1 = 4 \, \text{s} \) (2 seconds before the second ball is thrown and 2 seconds after).

2. **Calculating the Distance Covered by the First Ball:**
- Using the equation of motion:
\[ s_1 = u_1 t_1 + \frac{1}{2} a t_1^2 \]
- Here, \( a = g = 10 \, \text{m/s}^2 \) (acceleration due to gravity).
- Substituting the values:
\[ s_1 = 5 \cdot 4 + \frac{1}{2} \cdot 10 \cdot (4^2) \]
\[ s_1 = 20 + \frac{1}{2} \cdot 10 \cdot 16 \]
\[ s_1 = 20 + 80 = 100 \, \text{m} \]

3. **Understanding the Motion of the Second Ball:**
- The second ball is thrown downwards after 2 seconds, so it will be in motion for \( t_2 = 2 \, \text{s} \).
- Let the initial velocity of the second ball be \( u_2 \).

4. **Calculating the Distance Covered by the Second Ball:**
- Using the same equation of motion:
\[ s_2 = u_2 t_2 + \frac{1}{2} a t_2^2 \]
- Substituting the values:
\[ s_2 = u_2 \cdot 2 + \frac{1}{2} \cdot 10 \cdot (2^2) \]
\[ s_2 = 2u_2 + \frac{1}{2} \cdot 10 \cdot 4 \]
\[ s_2 = 2u_2 + 20 \]

5. **Setting the Distances Equal for Collision:**
- Since both balls collide, the distances covered by both balls must be equal:
\[ s_1 = s_2 \]
\[ 100 = 2u_2 + 20 \]

6. **Solving for \( u_2 \):**
- Rearranging the equation:
\[ 2u_2 = 100 - 20 \]
\[ 2u_2 = 80 \]
\[ u_2 = \frac{80}{2} = 40 \, \text{m/s} \]

### Final Answer:
The velocity with which the second ball should be thrown downwards is \( 40 \, \text{m/s} \).

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