1. Set up the motion of ball A

• Height of tower = $180\,\text{m}$
• They meet at $100\,\text{m}$ above ground.

So, distance fallen by each ball till meeting point is $180 - 100 = 80\,\text{m}.$

Ball A is released from rest at $t=0$.

Using $s = \frac{1}{2}gt^2,$ we get $80 = \frac{1}{2}(10)t^2 = 5t^2.$ So, $t^2 = 16 \Rightarrow t = 4\,\text{s}.$

Thus, ball A reaches the meeting point after $4\,\text{s}$ from its release.

2. Time of motion of ball B

Ball B is thrown downward at $t=2\,\text{s}$.

Since the meeting occurs at $t=4\,\text{s}$ from the start, ball B has been moving for $4 - 2 = 2\,\text{s}.$

3. Apply equation of motion to ball B

Ball B also covers $80\,\text{m}$ in $2\,\text{s}$ with initial downward velocity $u$.

Using $s = ut + \frac{1}{2}gt^2,$ $80 = u(2) + \frac{1}{2}(10)(2^2).$

So, $80 = 2u + 20.$ Therefore, $2u = 60 \Rightarrow u = 30\,\text{m/s}.$

4. Check options

• A: $10$ ❌
• B: $15$ ❌
• C: $20$ ❌
• D: $30$ ✅

Hence, the correct option is D.