From the top of a tower, a ball is thrown vertically upward which reaches the ground in 6 s. A second ball thrown vertically downward from the same position with the same speed reaches the ground in 1.5 s. A third ball released, from the rest from the same location, will reach the ground in _______s.

Let h be the height of tower. Situation for the first condition is shown in the figure.                   

By using equation of motion, $h=-u{t}_1+\frac{1}{2}g{t}_1^2;h=-6u+\frac{1}{2}g{\left(6\right)}^2$

h = - 6u + 18g ……..(i)

Situation for the second ‘s condition is shown in the figure.
              
Now, using equation of motion.
$\begin{array}{l}h=u{t}_2+\frac{1}{2}g{t}_2^2;h=1.5u+\frac{1}{2}g{\left(1.5\right)}^2\\ h=1.5u+\frac{2.25}{2}g\mathrm{..............}\left(ii\right)\end{array}$
Solving Eqs.(i) and  (ii) we get h = 4.5 gm For the third case, when ball is released from rest, i.e  u = 0
               
$\begin{array}{l}Inthiscase,h=0X{t}_3+\frac{1}{2}g{t}_3^2=\frac{1}{2}g{t}_3^2\\ \Rightarrow t_3=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2X4.5g}{g}}=\sqrt{9}=3s\end{array}$