A Sniper's job

Skip to the next paragraph for the question. This question is show how hard a sniper's job is and it is not like Hollywood where it is "one shot one kill". I am sure that question is not very intuitive but still try what a sniper calculates before a shot.

Imagine a sniper is on an island which is along the equator and is on top of hill of height 1000m, he is there to shoot his target who is at height of 0 m and the angle of depression of his target is $\alpha$ ( tan $\alpha$ = 2). He finds that the wind is blowing with a speed of 50 m/s along + z axis. He finds that his target will move at speed of 30 m/s along + x axis (the target moves immediately after the sniper fires). The sniper takes aim, 2 seconds after firing ( along the equator ) it hits the target.

Now, my question is that before the sniper fires his position vector of his sniper rifle makes the angles as shown in the above image. Let the speed of bullet be v. Find $\left \lfloor v × \frac{\sin\beta}{\sin\gamma} \right \rfloor$ - 3?

Note:

• Neglect air drag.
• Do not neglect Coriolis effect because sniper's don't.
• The ambient gravitational acceleration is 10 $\text{m/s}^2$ in the - $y$ direction.