A ball is dropped from a height. If it requires 0.200 s to cross...

Let the velocity of the ball at the beginning of the last $6.00 \text{ m}$ be $u$.

Using the equation of motion for the last $6.00 \text{ m}$:

$s = ut + \dfrac{1}{2}gt^2$

$6.00 = u(0.200) + \dfrac{1}{2}(10)(0.200)^2$

$6.00 = 0.200u + 5(0.04)$

$6.00 = 0.200u + 0.20$

$0.200u = 5.80$

$u = 29 \text{ m/s}$

This velocity is acquired by the ball after falling a height $h_1$ from rest.

Using $v^2 = u^2 + 2gh_1$ where initial velocity is zero:

$(29)^2 = 0 + 2(10)h_1$

$841 = 20h_1$

$h_1 = 42.05 \text{ m}$

The total height from which the ball was dropped is $H = h_1 + 6.00$.

$H = 42.05 + 6.00 = 48.05 \text{ m}$