A ball is thrown vertically upwards with a velocity of 19 . 6m s ^...

Initial velocity of ball is $u = 19 . 6m{s}^{- 1}$, final velocity $v = 0$.

Using $v = u + at$, here, acceleration $a = - g$.

Time taken in upward motion above tower is ${t}_{a} = \dfrac{u}{g} = \dfrac{19 . 6}{9 . 8} = 2s$

Now, time taken from top most point to ground is ${t}_{d} = 6 - 2 = 4s$.

Or, {t}_{d} = \sqrt{\dfrac{2{h}_\max}{g}} (Using $s = ut + \dfrac{1}{2}g{t}^{2}$)

⇒ {h}_\max = \dfrac{16 \times 9 . 8}{2} = \dfrac{392}{5}m.

Hence, the value of $k = 392$.