A ball is droped from a high rise platform `t = 0` starting from rest. After `6 s` another ball is thrown downwards from the same platform with a speed `v`. The two balls meet at `t = 18 s`. What is the value of `v` ?

Options

A. `74 ms^(-1)`
B. `55 ms^(-1)`
C. `40 ms^(-1)`
D. `60 ms^(-1)`

Answer

To solve the problem, we will analyze the motion of both balls separately and then set up an equation based on their positions when they meet.

### Step-by-Step Solution:

1. **Identify the Motion of the First Ball:**
- The first ball is dropped from rest at \( t = 0 \).
- It falls freely under gravity for \( 18 \) seconds.
- The distance fallen by the first ball, \( S_1 \), can be calculated using the equation of motion:
\[ S_1 = u_1 t + \frac{1}{2} g t^2 \]
where \( u_1 = 0 \) (initial velocity) and \( g \approx 9.8 \, \text{m/s}^2 \).
\[ S_1 = 0 \cdot 18 + \frac{1}{2} \cdot 9.8 \cdot (18)^2 \]
\[ S_1 = \frac{1}{2} \cdot 9.8 \cdot 324 = 1587.6 \, \text{m} \]

2. **Identify the Motion of the Second Ball:**
- The second ball is thrown downwards \( 6 \) seconds after the first ball, so it is in motion for \( 12 \) seconds when they meet.
- Let the speed of the second ball be \( v \).
- The distance fallen by the second ball, \( S_2 \), is given by:
\[ S_2 = v t + \frac{1}{2} g t^2 \]
where \( t = 12 \) seconds.
\[ S_2 = v \cdot 12 + \frac{1}{2} \cdot 9.8 \cdot (12)^2 \]
\[ S_2 = 12v + \frac{1}{2} \cdot 9.8 \cdot 144 = 12v + 705.6 \, \text{m} \]

3. **Set the Distances Equal:**
- Since both balls meet at the same point, we can set \( S_1 = S_2 \):
\[ 1587.6 = 12v + 705.6 \]

4. **Solve for \( v \):**
- Rearranging the equation gives:
\[ 12v = 1587.6 - 705.6 \]
\[ 12v = 882 \]
\[ v = \frac{882}{12} = 73.5 \, \text{m/s} \]

### Final Answer:
The value of \( v \) is \( 73.5 \, \text{m/s} \).