A body is projected horizontally with a velocity of $4\sqrt 2 {\text{m/s}}$.The velocity of body after $0.7$seconds will be nearly: (Take $g = 10{\text{m/se}}{{\text{c}}^2}$)
(A) $10{\text{m/sec}}$
(B) $9{\text{m/sec}}$
(C) $19{\text{m/sec}}$
(D) $11{\text{m/sec}}$

Hint
In a projectile motion there are 2 components of velocity with which the object moves, horizontal and vertical. In the path, the vertical velocity changes as the object first decelerates when thrown upwards and accelerates again in the downward direction. The horizontal velocity, on the other hand, remains the same.
$\Rightarrow v = u + at$ (first equation of motion)
Here v is the final velocity
u is the initial velocity of the object
a is the acceleration
and t is the time taken by it.
$\Rightarrow {\vec v_x} + {\vec v_y} = \sqrt {{{({v_x})}^2} + {{({v_y})}^2}} $ (Vector sum of 2 perpendicular components)

Complete step by step answer
In the question, it is told that the object is projected horizontally, this means, the object is thrown sideways. On analysing it’s horizontal and vertical components, we find that-
-Horizontal: It moves with a constant velocity of $4\sqrt 2 {\text{m/s}}$ which does change until the ball reaches ground.
-Vertical: It starts with zero upward acceleration, which means the only acceleration taking place throughout the flight will be downwards, at a constant rate of $g = 10m/s$.
After 0.7 seconds,
Horizontal velocity, ${v_x} = 4\sqrt 2 m/s$
Vertical velocity ${v_y} = {u_y} + gt$
Since it started with zero velocity, initial vertical velocity ${u_v} = 0$
Putting the other values as
$\Rightarrow g = 10m/s$
$\Rightarrow t = 0.7\sec $
We get,
$\Rightarrow {v_y} = 0 + 10 \times 0.7$
$\Rightarrow {v_y} = 7m/\sec $
For the total velocity after$0.7\sec $, we square and add both the horizontal and vertical components-
$\Rightarrow v = {\vec v_x} + {\vec v_y}$
$\Rightarrow v = \sqrt {{{(4\sqrt 2 )}^2} + {7^2}} $
$\Rightarrow v = \sqrt {32 + 49} $
$\Rightarrow v = \sqrt {81} $
$\Rightarrow v = 9{\text{ m/sec}}$
Thus option (B) is the correct answer.

Note
It is important to pay attention to initial conditions in such problems of projectile motion, especially about the angle of projectile. Also, the horizontal velocity in a projectile remains constant, unless stated otherwise, because it does not experience any force, thus no horizontal acceleration takes place.