A stone thrown vertically upward from a certain height. It covers half of distance covered during the time of flight to last second. The time of flight (in second) of the stone is__________

From figure,

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}v=v_0-g{t}_1\\ \text{ Or }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0=v_0-g{t}_1\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{t}_1=\frac{v_0}{g} ........\left(i\right)\\ \text{ and }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{v}^2=v_0^2-2gh\\ \text{ or }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{0}^2=v_0^2-2gh\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}h=\frac{v_0^2}{2g} .......\left(ii\right)\end{array}$

 

 

According to the question,

$h+H-\left(\frac{2h+H}{2}\right)=\frac{1}{2}g{\left(t_2-1\right)}^2 ......\left(iv\right)$

 

From Eqs. (i), (ii), (ii) and (iv), we get

Time of flight = t₁ + t₂ =4s