A stone is thrown vertically upward with an initial velocity $v_o.$ The distance travelled in time $\frac{1.5v_o}{g}$ is

The correct option is C $\frac{5v_o^2}{8g}$
Velocity of particle will become zero in time,
$t_o=\frac{v_o}{g}$ (by using first equation of motion)
The given time $t=\frac{1.5v_o}{g}$ is greater than the time $t_o=\frac{v_o}{g}$
Hence, The total distance travelled will be the sum of the distance travelled in time $t_o$ and $(t-t_o)$
Distance travelled $=s_t+s_{t-t_o}$
By using third equation of motion, $0^2=v_o^2-2g{s}_t$
$s_t=\frac{v_o^2}{2g}$
For $s_{t-t_0}$, using second equation of motion, $s_{t-t_o}=(0\times t)+\frac{1}{2}g(t-t_o{)}^2$
$s_{t-t_o}=\frac{1}{2}g{(\frac{1.5v_o}{g}-\frac{v_o}{g})}^2$
$s_{t-t_o}=\frac{1}{2g}(.5v_o{)}^2$
$s_{t-t_o}=\frac{v_o^2}{8g}$
Distance travelled $=\frac{v_o^2}{2g}+\frac{v_o^2}{8g}\Rightarrow \frac{5v_o^2}{8g}$