For an "exact" solution, we can also use the following variant of the argument of Henning Makholm.
Let the height above the window be $h$, let the total time to reach the top of the window be $t_1$, and total time to reach the bottom of the window be $t_2$. Let acceleration due to gravity be constant at $a=9.8$.
Then
$$h=\frac{1}{2}at_1^2 \qquad\text{and}\qquad h+1.35=\frac{1}{2}at_2^2.$$
Subtract. We get
$$\frac{1}{2}a(t_2^2-t_1^2)=1.35.$$
But $t_2^2-t_1^2=(t_2-t_1)(t_2+t_1)$. Since $t_2-t_1=0.210$ we obtain
$$t_2+t_1=\frac{(2)(1.35)}{(0.210)(9.8)}.$$
Note that $t_1=(1/2)((t_2+t_1)-(t_2-t_1))$. Use our expression for $t_2+t_1$, together with $t_2-t_1=0.210$, to find $t_1$. Now we know $t_1$, so we know $h$.
Another way: It is more pleasant to avoid numbers until the end. Let $w$ be the height of the window, and let $s$ be the amount of time it took for the object to pass the window. Let $u$ be the velocity at the top of the window, and
$v$ the velocity at the bottom. Then the average velocity at which the window was traversed is $(v+u)/2$. But it is also $w/s$, and therefore
$$\frac{v+u}{2}=\frac{w}{s}.$$
The change in velocity is $v-u$. It is also $as$. Thus
$$\frac{v-u}{2}=\frac{as}{2}.$$
From the above two equations it follows that
$$u=\frac{w}{s}-\frac{as}{2}.$$
The average velocity from the time of dropping until arriving at the window is $u/2$. The time it took is $u/a$, so the distance travelled is $u^2/2a$. It follows that the height above the window from which the object was dropped is
$$\frac{\left(\frac{w}{s}-\frac{as}{2}\right)^2}{2a}.$$
You need to look up something called Snell's law and critical angle. If you are directly over your target, there is no refraction since the angle of incidence $\theta_i$, your divergence from perpendicular to the surface of the water, and the angle or refraction $\theta_r$, the object's divergence from perpendicular to the surface of the water, are both zero. When you are approaching your target, say in a boat, the relationship between $\theta_i$ and $\theta_r$ is Snell's Law:
$$
\frac{n_1}{n_2}=\frac{\sin{\theta_i}}{\sin{\theta_r}}
$$
where $n_1$ is the index of refraction of air (approximately equal to $1$) and $n_2$ is the index of refraction of water ($1.33$). These indices are actually measures of how much the speed of light slows down in the absence of a vacuum, such that $n=\frac{c}{v}$ where $c$ is the speed of light.
The difference between the angle of incidence and the angle of refraction makes an object under the water appear to be closer to the surface than it actually is. Suppose you are a spear-fisherman, your point of view is very close to the surface of the water, and your object is to skewer a large fish a meter or more underneath the water with an accurate throw.
Starting with your spear pointing straight down at the water, you would raise it an initial angle $\theta_i$ and then lower it by an amount $\theta_i-\sin^{-1}{(1.33\sin{\theta_i})}$ to hit your target. If you are looking at the water from a height which is within an order of magnitude to the depth of your target, the math gets much more complicated. It is somewhat indicative of the power of the human brain that fishmen in primitive cultures, with no formal knowledge of the effects of light refraction, are exceedingly good at this.
When the angle of refraction becomes very large, which happens as depth increases, the angle of incidence may exceed $90\,^\circ$. This is called the critical angle and it results in a total internal reflection of light under the water. Thus, as a target gets deeper into the water, the more nearly perpendicular your point of vantage needs to be. Suppose you have two fish, one at a depth of $1\,\rm m$ and the other directly under it at a depth of $2\,\rm m$. As you approach, you will see the shallower fish first, and even though the water may be glassy smooth and crystal clear the fish at lower depths will still be invisible to you until you get closer.
Light attenuation is the amount of light left after it has passed through a certain distance of absorbing media. For a complete description, you will want to look up the Beer-Lambert law. I'll give you the highlights. Liquid water quickly absorbs most wavelengths of electromagnetic radiation, but readily admits light in the $360-750\,\rm nm$ range (visible light). As you go deeper, the amount of available light undergoes an exponential decay. Even so, after $200\,\rm m$ there is virtually no transmission of light.
The longer wavelengths (red) not only deviate the most under diffraction but also are attenuated quickest, while the shorter (blue-violet) wavelengths diffract the least and penetrate the most (this is why the deep ocean appears blue and deep-sea creatures are so often red-colored: makes them invisible). The more light diffracts, the greater the distance it has to travel through a medium to reach your eye. This, combined with light intensity's exponential decay with respect to distance is what makes deep objects "disappear" so quickly.
So the answer to your question is YES, there is a relationship between refraction and attenuation. Although the two effects are separate, they work together to make red-colored items more difficult to see at great underwater distances. Also, you can see the deepest when you look straight down into the water, as previously noted.
Incidentally, if you've ever been scuba-diving or snorkeling, you may have noticed that there are "shafts of light" in the water. This is a good example of what you are talking about. Non-vertical light diffracts the most and is quickly attenuated, so the greatest penetration of light occurs in vertical shafts, an effect you also notice while looking up at moisture-laden rain clouds. (For me, understanding how this effect works only adds to their beauty.)
Best Answer
The question shows you already found the correct numeric result. I would just suggest working on the presentation so that people reading your calculations do not misunderstand them. (Eventually you may also be solving problems that are complicated enough that even you will not be able to remember what you're doing unless you explain it carefully.)
I would assign a different number or name to the time of each event. It often helps to choose a convenient event and measure all times relative to that event. So if the event is when the first drop starts falling, the first drop starts falling at $0$ seconds and the second drop starts falling at $1$ second. When you are first setting up the problem you have not yet calculated when the first drop hits the ground, so give that a name: you might say it happens at $t_1$ seconds. And you can say the second drop hits the ground at $t_2$ seconds.
So the first drop falls for a total of $t_1$ seconds, which gives a vertical distance of $$ 0=1000−5t_1^2.$$ From this is follows that $t_1 = 10\sqrt2$; this is really the same thing you already did, just giving a unique name to this time so it's clear which time we're talking about.
At time $t_1$, when the first drop hits the ground, the second drop has only been falling for $t_1 - 1$ seconds. So its height above the ground at that instant is $$1000 - 5(t_1 - 1)^2 = 1000 - 5(10\sqrt2 - 1)^2 \approx 136.$$
And of course at $t_2$ seconds, when the second drop hits the ground, the drop has been falling for $t_2 - 1$ seconds, and so $$ 0=1000−5(t_2 - 1)^2.$$ From this we can figure out that $t_2 = 10\sqrt2 + 1$; personally, however, I would just observe that since the two drops travel the same distance with the same initial velocity and acceleration, they take the same elapsed time, so $t_2 - 1 = t_1 = 10\sqrt2$ and we can immediately solve this to find that $t_2 = 10\sqrt2 + 1 = t_1 + 1$ without dealing with a quadratic.
If we wanted a more long-winded solution I suppose we could say that if the second drop has been falling for $\Delta t_{21}$ seconds when the first drop hits the ground, then $t_1 = \Delta t_{21} + 1$ (because the first drop started one second earlier) and therefore $$ 0=1000−5(\Delta t_{21} + 1)^2.$$ That's the equation $0=1000−5(t + 1)^2$ that you wrote, but writing $\Delta t_{21}$ instead of $t$ is a reminder that this equation was true with regard to a very specific amount of elapsed time. we can solve that equation to find out that $\Delta t_{21} = 10\sqrt2 - 1$, and then find the height of the second drop above the ground at time $t_1$ by evaluating $1000 - 5(\Delta t_{21} - 1)^2,$ but I'm already having more trouble remembering which symbol means what than I did in the previous solution, so I think maybe this isn't the easiest way to go.