A ball is thrown vertically upwards with a velocity of $19.6 \mathrm{m} {\mathrm{s}}^{-1}$ from the top of a tower. The ball strikes the ground after $6 \mathrm{s}$. The height from the ground up to which the ball can rise will be $\left(\frac{\mathrm{k}}{5}\right) \mathrm{m}$. The value of $k$ is _____ (use $\mathrm{g}=9.8 \mathrm{m} {\mathrm{s}}^{-2}$)

Initial velocity of ball is $u=19.6 \mathrm{m} {\mathrm{s}}^{-1}$, final velocity $v=0$.

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

Time taken in upward motion above tower is $t_a=\frac{u}{g}=\frac{19.6}{9.8}=2 \mathrm{s}$

Now, time taken from top most point to ground is $t_d=6-2=4 \mathrm{s}$.

Or, $t_d=\sqrt{\frac{2h_{\max }}{g}}$    (Using $s=ut+\frac{1}{2}g{t}^2$)

$\Rightarrow h_{\max }=\frac{16\times 9.8}{2}=\frac{392}{5} \mathrm{m}$.

Hence, the value of $k=392$.