Question

A ball is thrown vertically upwards from top of a tower of height H with velocity V the ball strikes the ground after time

Answer 1

 i think time taken would be v/9.8

Answer 2

The height is given by,

                $h = -vt+\frac { 1 } { 2 } g { t }^{ 2}$

                $g{ t }^{ 2}- 2vt- h= 0$

                $t = \frac{-(-2v)\pm \sqrt { 4 { v }^{ 2 }+ 4gh } } { 2g }$

                $t = \frac { 2v\pm 2\sqrt { { v }^{ 2 }+ gh } } { 2g }$

                $t = \frac { v } { g } \pm \frac { { \left[{v }^{ 2 }+ 2gh \right] }^{ 2} } { g }$

                $t = \frac { v } { g } \left[ 1+ \sqrt { 1+ \frac { 2gh } { { v }^{ 2 } } }\right]$