Two tall buildings  are 30 m apart.   The speed  with which a ball must  be thrown horizontally  from a  window 150 m above the ground in  one  building so that it enters a  window 27.5m from  the ground  in the other building is

Since the ball is thrown horizontally, it travels  with uniform velocity in the horizontal  direction while it also accelerates in vertical  direction due to gravity
Given :
Distance travelled in horizontal direction =  30m
Distance travelled in vertical direction (150- 27.5) m
Assumption :
Horizontal velocity $\left(\overrightarrow{v}\right)=\mathrm{u}$
Total time taken = t
$g=9.8m{s}^{-2}$
Now, Horizontal  distance = ut (speed x time )
Ut = 30 _____ (i)
Vertical distance $=\frac{1}{2}g{t}^2$
$\frac{1}{2}g{t}^2=122.5$
Put the value of t from equation (i) to Eq. 
(ii) $\begin{array}{r}\frac{1}{2}\left(9.8\right){\left(\frac{30}{4}\right)}^2=122.5\\ u^2=\left[\frac{9.8\times 30\times 30}{2\times 122.5}\right]\\ u=\sqrt{6}=6\mathrm{m}/\mathrm{s}\end{array}$
Hence option B is correct.