From the top of a tower, a ball is thrown vertically upward which reaches the ground in 6   s . A…

When it is thrown upward, $-h=v\times 6-\frac{1}{2}g{\left(6\right)}^2$

When it is thrown downward, $-h=-v\times 1.5-\frac{1}{2}g{\left(1.5\right)}^2$

Therefore, $5h=5\left[36+9\right]$ $h=45 \mathrm{m}$

When the ball is dropped, $h=\frac{1}{2}g{t}^2\Rightarrow 45=\frac{1}{2}\times 10\times t^2\Rightarrow t=3 \mathrm{s}$

Alternate method, $t=\sqrt{t_1{t}_2}=\sqrt{6\times 1.5}=3 \mathrm{s}$