A stone is thrown vertically upward with an initial velocity $v_o$. The distance travelled in time $\frac{4v_o}{3g}$ is

The correct option is D $\frac{5v_o^2}{9g}$
To calculate the total distance time taken to go vertically up untill the velocity becomes zero and time taken to come downward should be considered.
Let the time taken by stone to go vertically up be $t_1$, using $1^{st}$ equation of motion,
$v=u-g{t}_1$
$0=v_o-g{t}_1$
$\Rightarrow t_1=\frac{v_o}{g}$
Hence time taken to come downward $t_2$ $=\frac{4v_o}{3g}-\frac{v_o}{g}=\frac{v_o}{3g}$
Distance covered vertically upward, from second equation of motion,
$s_1=u{t}_1-\frac{1}{2}g{t}_1^2$ (downward negative)
Putting the value of $t_1=\frac{v_o}{g}$ and $u=v_o$,
$s_1=\frac{v_o^2}{g}-\frac{1}{2}g{\left(\frac{v_o}{g}\right)}^2$
$\begin{array}{}{s}_1=\frac{v_o^2}{g}-\frac{v_o^2}{2g}=\frac{v_0^2}{2g}........& \text{(1)}\end{array}$
Similarly, distance covered vertically upward, from second equation of motion,
$s_2=u{t}_2-\frac{1}{2}g{t}_2^2$ (downward negative)
Putting the value of $t_2=\frac{v_o}{3g}$ and $u=0$,
$s_2=0-\frac{1}{2}g{\left(\frac{v_o}{3g}\right)}^2=\frac{v_o^2}{18g}$
$\begin{array}{}{s}_2=\frac{v_0^2}{18g}....& \text{(2)}\end{array}$
Hence total distance covered, $s=s_1+s_2=\frac{v_0^2}{2g}+\frac{v_0^2}{18g}=\frac{5v_0^2}{9g}$