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 m//s`
B. `64 m//s`
C. `84 m//s`
D. `94 m//s`

Explanation

To solve the problem, we need to analyze the motion of both balls and find the value of the speed \( v \) of the second ball thrown downwards.

### Step-by-step Solution:

1. **Identify the Motion of the First Ball:**
- The first ball is dropped from rest at \( t = 0 \).
- The time of travel for the first ball when they meet is \( t = 18 \) seconds.
- The distance traveled by the first ball can be calculated using the equation of motion:
\[ S_1 = ut + \frac{1}{2} g t^2 \]
Here, \( u = 0 \) (initial velocity), \( g \approx 9.8 \, \text{m/s}^2 \), and \( t = 18 \, \text{s} \).
\[ S_1 = 0 + \frac{1}{2} \cdot 9.8 \cdot (18)^2 \]
\[ S_1 = \frac{1}{2} \cdot 9.8 \cdot 324 = 1587.6 \, \text{m} \]

2. **Identify the Motion of the Second Ball:**
- The second ball is thrown downwards after 6 seconds, so it travels for \( t = 18 - 6 = 12 \) seconds.
- The distance traveled by the second ball is given by:
\[ S_2 = vt + \frac{1}{2} g t^2 \]
Here, \( t = 12 \, \text{s} \).
\[ S_2 = v \cdot 12 + \frac{1}{2} \cdot 9.8 \cdot (12)^2 \]
\[ S_2 = 12v + \frac{1}{2} \cdot 9.8 \cdot 144 = 12v + 705.6 \, \text{m} \]

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

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

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