Question

The initial speed of a bullet fired from a rifle is $630m/s$. The rifle is fired at the centre of a target $700m$ away at the same level as the target. How far above the centre of the target rifle must be aimed in order to hit the target? (Take$g=10m/{{s}^{2}}$)
A. $9.8m$
B. $4.2m$
C. $1.0m$
D. $6.17m$

Answer 1

Here initial velocity, distance from the target and gravity are given. We need to find the position at which centre of the target rifle must be aimed in order to hit the target. When the bullet is fired, it will have a projectile motion. We have learned the relation between distance covered, velocity and gravity of projectile motion. By using this horizontal range equation, we can solve the question.
Formula used:
$h=ut+\dfrac{1}{2}g{{t}^{2}}$
$velocity, v=\dfrac{displacement}{time}$

Complete answer:
Given,
$\text{Initial velocity of bullet = 630 m/s}$
$\text{Distance from the center of target = 700 m}$
$g=10m/{{s}^{2}}$
Let’s find the solution using vertical height equation,
$h=ut+\dfrac{1}{2}g{{t}^{2}}$------ 1
Where,
$\text{u - initial velocity}$
$\text{h - vertical height}$
$\text{g- acceleration due to gravity}$
$\text{t- time taken for the vertical fight}$
For vertical motion of the bullet,
$u=0$------- 2
Let, $t$ be the time taken by the bullet to hit the target with velocity $v=630m/s$, and displacement, $d=700m$. Then,
$time=\dfrac{displacement}{velocity}=\dfrac{700}{630}=\dfrac{10}{9}s$ -------- 3
Substitute 2 and 3 in equation 1. We get,
$h=\dfrac{1}{2}\times 10\times {{\left( \dfrac{10}{9} \right)}^{2}}=\dfrac{500}{81}=6.1m$
Therefore, the bullet should be fired $6.17m$ above the center of target.

So, the correct answer is “Option D”.

Additional Information:
According to Newton's first law of motion a bullet fired from a gun will continue in motion at the same speed unless an external force is acted on it. The horizontal motion of the bullet is constant. Vertically, the bullet is uniformly accelerated by gravity. Ignoring air resistance, the only force acting on a bullet is gravity. The downward motion of a bullet is the same as a body in free-fall. Hence bullets fired from a rifle have a projectile path.

Note:
We have,
$h=ut+\dfrac{1}{2}g{{t}^{2}}$
For a vertical motion of the bullet,
$u=0$
Then,
$t=\sqrt{\dfrac{2h}{g}}$---- a
Given,
$h=700m$
$g=10m/{{s}^{2}}$
We know,
$velocity, v=\dfrac{displacement}{time}=\dfrac{d}{t}$
Then,
$d=vt$--- b
Substitute equation a in b. We get,
$d=\sqrt{\dfrac{2h}{g}}\times v$ ---- c
Substituting the values of $\text{d,g and v}$in equation c, we get,
$700=\sqrt{\dfrac{2\times h}{0}}\times 630$
Then,
$h=6.17m$