Question

The muzzle velocity of a certain rifle is \[330\;m{s^{ - 1}}\] at the end of one second, a bullet fired straight up into the air will travel a distance of.
A. \[\left( {330 - 4.9} \right)m\]
B. \[\left( {330 + 4.9} \right)m\]
C. \[330m\]
D. \[\left( {330 - 9.8} \right)m\]

Answer 1

Hint: We need to know basic laws of motion
\[
  v = u + at \\
  {v^2} = {u^2} + 2gs \\
  s = ut + \dfrac{1}{2}g{t^2} \\
 \]

We need to take the equation that has time, distance and velocity which is given in the question

Complete step by step answer:
Muzzle velocity is the velocity with which a bullet or shell leaves the muzzle of a gun, that is this is the speed of a projectile with respect to the muzzle at the moment it teases the end of the gun's barrel. A higher muzzle vilify correlates to a flatter trajectory, meaning there is less bullet drop or a given distance when the velocity of a projectile is increased.

We know
$s = ut + \dfrac{1}{2}g{t^2}$

Where s is the distance travelled, u is the initial velocity, t is the time of protection, g is the acceleration due to gravity which is negative when a body moves up and if a body is moving downwards it will be positive.
So,
$s = ut - \dfrac{1}{2}g{t^2}$
By the problems
$
  u = 330\dfrac{m}{s} \\
  t = 1s \\
  g = 9.8\dfrac{m}{{{s^2}}} \\
  \therefore s = (330 \times 1) - \dfrac{1}{2} \times 9.8 \times {1^2} \\
  s = (330 - 4.9)m \\
 $

So, the correct answer is “Option A”.

Note:
We should take care of g, g is the acceleration due to gravity which is negative when a body moves up and if a body is moving downwards it will be positive.
When a gun is fired this takes a projectile mutation in general. But here is not given to write any degrees of projection. I have the problem that on the basis of projectile motion a muzzle velocity correlates to a flatter meaning there is less velocity drop over a given distance when the velocity of a projectile is increased.