Given, distance between the two buildings d=100 m height of each tower, h=200 m speed of each bullet, v=25 ms−1 The situation can be shown as below 
where, x be the vertical distance travelled from the top of the building and t be the time at which they collide. As two bullets are fired toward each other, So, their relative velocity will be vrel =25−(−25)=50 ms−1 Then, time t=vrel d=50100=2 s The distance or height at which they collide is calculated from equation of motion, x=ut+21at2 The bullet is initially a rest i.e. u=0 and as it is moving under the effect of gravity a=−g, so x=−21g2x=−21×10(2)2=−20 m The negative sign shows that the bullets will collide 20 m below the top of tower i.e. at a height of (200−20)=180 m from the ground after 2 s.