A classical mechanics problem by Swapnil Das

Here, $h=6m$

As the acceleration due to gravity is not mentioned, we take $g=10m/s^{2}$

Let's first find the initial velocity, when the object is 6 m above the ground.

Using $h=ut+ \dfrac{gt^{2}}{2}$ , we get,

$6 = 0.2u+\dfrac{10\times 0.2^{2}}{2}$

$6=0.2u+0.2$

Or, $u=\dfrac{5.8}{0.2}=\boxed{29m/s}$ .

The initial velocity we've found earlier, is the final velocity of the journey above. Therefore, $v=29m/s$ .

And the initial velocity i.e, $u=0m/s$

Therefore, the distance traveled by the stone before reaching 6 m above the ground is given by $h=\dfrac{v^{2}-u^{2}}{2g}$

By substituting the values, $h= \dfrac{29^{2}-0^{2}}{2\times 10}$

$h=\dfrac{29\times 29}{20}$

Or, $h=42.05 \approx \boxed{42}$

Therefore the total height = $42+6=\boxed{\boxed {\boxed{ 48 }}}$