Apoliceman is chasing a criminal across a rooftop at 10 m/s. he decides to jump to the next building which is 2 meters across from the original and 2.5 meters lower. how far (in m) will they fall in that time? (use g = 10 m/s2)

Reading the scenario, we can say that the motion of the policeman follows a horizontal trajectory.  The motion is both vertical and horizontal so we need to consider these components separately and at the same time, together. 

What we are looking for is the horizontal distance (dx). 

You can derive all your formulas from one formula:


Where:
d = displacement
Vi = initial velocity
t = time in flight
g = Acceleration due to gravity

When using this formula, you can use only horizontal components if you are looking for a value of the horizontal movement. If you are looking for vertical components you use only values of the vertical movement. 

In projectiles, horizontal motion is constant and gravity is not an issue. So the value of g is going to be 0, which will leave you with the formula:



Vix = initial vertical horizontal velocity
t = time in flight

Notice we do not have time in this case.  But we do have vertical height, which we can use to get the time of flight. Viy is 0m/s because, in the beginning, the object is not moving vertically yet. So you will be left with:



Where:
dy = vertical height

From there we derive a formula for time and you will have:



So let's just plug in what we know from there:
Given:
dy = 2.5m
Vix = 10m/s
g = 10 m/s







So our time in flight is 1.41s.

Now that we have that we can use it in our horizontal motion, to get the horizontal distance. 





So they will fall 14.1 m FAR. 
If what you will need is how far from the edge of the next building, then just subtract the gap from the total distance:
14.1m - 2m = 12.1m

They will fall 12.1m away from the edge of the second building. 

I hope you understood my explanation. 

Initial velocity,  u = 4.9 m/sfinal velocity = v m/sinitial height,  h₁ =  245 m    final height, h₂ = 0a = -9.8  m/s²time taken = t secondss = ut + 0.5at²⇒  (h₂-h₁) = ut + 0.5at²⇒  0-245 = 4.9t + 0.5×(-9.8)×t²⇒ -245 = 4.9t - 4.9t²⇒ 4.9t² -4.9t -245 =0solving it, we get   t = 7.59sv  = u +  at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/svelocity is  69.48 m/s downward