Two balls $$A$$ and $$B$$ are placed at the top of 180 m tall tower. Ball $$A$$ is released from the top at $$t = 0$$ s. Ball $$B$$ is thrown vertically down with an initial velocity $$u$$ at $$t = 2$$ s. After a certain time, both balls meet 100 m above the ground. Find the value of $$u$$ in m s$$^{-1}$$. [use $$g = 10$$ m s$$^{-2}$$]

Ball A is released from the top of a 180 m tower at $$t = 0$$. Ball B is thrown down with velocity $$u$$ at $$t = 2$$ s. They meet 100 m above ground, i.e., 80 m below the top.

To find the time taken by Ball A to fall 80 m, we use $$s = \frac{1}{2}gt_A^2$$:

$$80 = \frac{1}{2}(10)t_A^2$$

$$t_A^2 = 16 \implies t_A = 4 \text{ s}$$

Since Ball B is thrown 2 s after Ball A, its travel time is

$$t_B = t_A - 2 = 4 - 2 = 2 \text{ s}$$

Next, substituting into $$s = ut_B + \frac{1}{2}gt_B^2$$ to find the initial velocity of Ball B gives

$$80 = u(2) + \frac{1}{2}(10)(4)$$

$$80 = 2u + 20$$

$$2u = 60 \implies u = 30 \text{ m s}^{-1}$$

The correct answer is Option D: 30.