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**Hint:**Always watch the sign convention when using Newton’s equations of motion. The acceleration is equal to acceleration due to gravity.

Formula Used: The formulae used in the solution are given here.

$S = ut + \dfrac{1}{2}a{t^2}$ where $S$ is the distance, $u$ is the initial velocity, $t$ is the time taken and $a$ is the acceleration, where $a = g$, acceleration due to gravity.

**Complete Step by Step Solution:**It has been given that, a rifle with a muzzle velocity $1500ft{s^{ - 1}}$ shoots a bullet at a small target 150ft away.

Muzzle velocity is the speed of a projectile (bullet, pellet, slug, ball/shots or shell) with respect to the muzzle at the moment it leaves the end of a gun's barrel (i.e. the muzzle). So, this is a case of projectile motion.

According to Newton’s law of motion, we have, $S = ut + \dfrac{1}{2}a{t^2}$ where $S$ is the distance, $u$ is the initial velocity, $t$ is the time taken and $a$ is the acceleration, where $a = g$, acceleration due to gravity.

The time taken to reach target by bullet = $t = \dfrac{{distance}}{{speed}} = \dfrac{{150}}{{1500}} = 0.1s$

Since we have motion in both the x-direction and the y-direction, write Newton’s equation in the y-direction since we are interested in how long the object is in the air. Use subscripts to help:

${S_{\left( y \right)}} = ut + \dfrac{1}{2}{a_y}{t^2}$.

Here, ${a_y} = g$. Thus, displacement along vertical (y) direction is given by,

${S_{\left( y \right)}} = 0 + \dfrac{1}{2} \times 9.8 \times {0.1^2}$ since

${S_{\left( y \right)}} = 0.049cm = 1.92inch$.

**Hence the correct answer is Option D.**

**Note:**It is assumed that the air resistance is negligible. Thus the result is only applicable when air resistance is zero. It is given that the air resistance is negligible, so the air resistance is left out from the equation.

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