From the top of a tall building, a gun is fired. The bullet leaves the gun at a speed of 340 m/s.?

Question

The bullet leaves the gun parallel to the ground and puts a hole in a window of another building and hits the wall that faces the windo. Using the data in the drawing, determine the distances D and H which locate the point where the gun was fired. Assume that the bullet does not slow down as it passes through the window.

Answer 1

Since the gun was fired parallel to the ground measure the hight from the floor to where it penetrated the window (h1) and then measure the height from the floor to where it hit the apposite wall (h2). (I hope the the floor is leveled well). next measure the distance (S) from the window to the wall.
The difference between the heights is the height (h) the bullet traveled in free fall due to gravity however with initial velocity

h=h1 - h2
Let t1 = time from the firing to hitting the window
t2 – time from hitting the window to hitting the wall

So h= V0 t2 + 0.5 g (t2)^2
The V0=g t1
Since horizontal velocity is constant (no air resistance)
t2=s/Vh

We have (substituting V0 and t2)
So h= g t1 s/Vh + 0.5 g (s/Vh)^2
Now
t1= [h - 0.5 g (s/Vh)^2] Vh/ ( s g)
Where g= 9.81 m/s^2

The horizontal distance from firing point to the window is
D=Vh t1
The height above the entry H
H=0.5 g (t1)^2

This is it.