From the top of a tower, a ball is thrown vertically upward which...

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

When it is thrown downward, $- h = - v \times 1 . 5 - \dfrac{1}{2}g{\left(1 . 5\right)}^{2}$

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

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

Alternate method, $t = \sqrt{{t}_{1}{t}_{2}} = \sqrt{6 \times 1 . 5} = 3s$