Question

A bullet is fired with a speed of $1500 m/s$ in order to hit a target $100 m$ away. If $g = 10\,{\text{m/}}{{\text{s}}^2}$, the gun should be aimed
A. $5 cm$ above the target
B. $10 cm$ above the target
C. $2.2 cm$ above the target
D. Directly towards the target

Answer 1

Use the kinematic equation which does not have final velocity term and calculate the time taken by the bullet to reach the target. Again, use the same kinematic equation but in the vertical motion of the bullet to calculate the vertical distance moved by the bullet. This is the distance that you should aim the gun above the target.

Formula used:
$s = ut + \dfrac{1}{2}a{t^2}$
Here, s is the distance, u is the initial velocity, a is the acceleration, and t is the time.

Complete step by step answer:
When we fire a bullet in order to hit the target placed at some distance, the bullet will eventually travel some vertical downward distance before it hits the target. This is due to acceleration due to gravity acting on the bullet to move downward. Let’s calculate the time taken by the bullet to reach the target using the following kinematic relation in the horizontal direction.
$s = {u_x}t + \dfrac{1}{2}{a_x}{t^2}$
Here, s is the distance of the target from the gun, ${u_x}$ is the horizontal component of the initial velocity, ${a_x}$ is the horizontal component of acceleration and t is the time taken.

Since there is no acceleration in the horizontal direction, we can write the above equation as,
$s = {u_x}t$
$\Rightarrow t = \dfrac{s}{{{u_x}}}$

Substituting 100 m for s and 1500 m/s for ${u_x}$ in the above equation, we get,
$t = \dfrac{{100}}{{1500}}$
$\Rightarrow t = 0.0667\,{\text{s}}$

Now, let’s use kinematic equation in the vertical motion of the bullet as follows,
${s_y} = {u_y}t + \dfrac{1}{2}g{t^2}$
Here, ${s_y}$ is the vertical distance traveled by the bullet and ${u_y}$ is the vertical component of initial velocity.

Since the bullet has only horizontal direction when it is fired, the vertical component of its initial velocity should be zero. Therefore, we can write the above equation as,
${s_y} = \dfrac{1}{2}g{t^2}$
Substituting $g = 10\,{\text{m/}}{{\text{s}}^2}$ and 0.0667 s for t in the above equation, we get,
${s_y} = \dfrac{1}{2}\left( {10} \right){\left( {0.0667} \right)^2}$
$\Rightarrow {s_y} = 0.022\,{\text{m}}$
$\therefore {s_y} = 2.2\,{\text{cm}}$
Thus, the bullet will move 2.2 cm in the downward direction. Therefore, we have to aim the gun 2.2 cm above the target.

Hence, option C is the correct answer.

Note: Note that the acceleration due to gravity is only along the vertical direction. Therefore, for the horizontal component of the motion, the term acceleration due to gravity is absent. Also, when we throw an object in the horizontal direction, initially it has only a horizontal component of velocity.