A rifle shoots a bullet with a muzzle velocity of 400 $m{s^{ - 1}}$at a small target 400m away. The height above the target at which the bullet must be aimed to hit the target is (g=10$m{s^{ - 2}}$).
A. 1m
B. 5 m
C. 10 m
D. 0.5 m

Hint: In order to explain this question we need to understand the concept of projectile motion. According to the question, when a bullet is fired it travels in projectile motion and then it deflects towards the earth due to the gravity of earth.

Complete step by step answer:
Let the velocity of the bullet along the X-axis and Y-axis which is given by
$
{V_x} = 400\cos \theta \\
\Rightarrow{V_y} = 400\sin \theta $
For small angle $\theta $
$
{V_x} = 400m{s^{ - 1}} \\
\Rightarrow{V_y} = 0 $
The time taken by the bullet is
$
t = \dfrac{{{V_x}}}{d} \\
\Rightarrow t = \dfrac{{400}}{{400}} \\
\Rightarrow t = 1\sec $
To calculate the height covered by the bullet to hit target is given by
 Vertical height above which the bullet must be aimed will be
$
y = {V_y}t + \dfrac{1}{2}g{t^2} \\
\Rightarrow{V_y} = 0 \\
\Rightarrow y = \dfrac{1}{2}g{t^2} \\
\Rightarrow y = \dfrac{1}{2} \times 10 \times {(1)^2} \\
\therefore y = 5m $
(This height will be the maximum height of the projectile motion)
Hence, the bullet must be aimed 5 m above the target so that it hits at the desired target.

Hence the correct option is B.

Note: Students should not get confused in horizontal velocity and vertical velocity while calculating the value of respective quantity. Also in the above question the target given is very small therefore we have considered the angle made by the bullet fired along the horizontal also be small.