Given diagram shows a railway compartment 16 m long, 3.2 m high and 2.4 m wide. If it is moving with a velocity v .A particle moving horizontally with a velocity of u, perpendicular to the velocity v, enters through a hole at an upper corner A and strikes the diagonally opposite corner B. which of the following value of v and u is (are) correct? Assume g =10 m/s
   

(A) v = 20 m/s
(B) u = 3 m/s
(C) v = 15 m/s
(D) v = 2 m/s

Hint There are three equations of motion that can be used to derive components such as displacement(s), velocity (initial and final), time (t) and acceleration (a). The following are the three equation of motion:
First Equation of Motion: $v = u + at$
Second Equation of Motion: $s = ut + \dfrac{1}{2}a{t^2}$
Third Equation of Motion: ${v^2} = {u^2} + 2as$

Complete step by step answer
Consider the vertical motion of the particle after entering the compartment. Let it reach the floor in time t. By using the second equation of motion.
$s = ut + \dfrac{1}{2}a{t^2}$ Here, s =$3.2m$ , a = $g = 10m/s$
Substituting values in the equation. (The u = 0 because the direction is perpendicular)
$3.2 = \dfrac{1}{2}\left( {10} \right){t^2}$
${t^2} = \dfrac{{3.2*2}}{{10}} = 0.64$
$t = 0.8\sec $.
Due to the velocity component u it covers $2.4m$ in this time.
Hence , $\
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\ $
$u = \dfrac{{2.4}}{{0.8}} = 3m/s$
Also , the distance moved by the train in this time is the length of the train with speed v
Hence , $v = \dfrac{{16}}{{0.8}} = 20m/s$

Hence, (a) and (b) options are correct.

Note At the highest point of a projectile the acceleration is perpendicular to velocity, and after that the acceleration is downwards velocity may increase or decrease depending on the object also same happens when there is upward motion of the object.