Question

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

Answer 1

$t _{ a }=\frac{ u }{ g }=\frac{19.6}{9.8}=2\, s$
$ t _{ d }=6-2 s =\sqrt{\frac{2 h _{\max }}{ g }}$
$ \Rightarrow h _{\max }=\frac{16 \times 9.8}{2}=\frac{392}{5}$