Question

A rifle is aimed at a target 40m away. The bullet hits the target 2.2cm below the horizontal plane (ignore wind and rotational effects).
(a) What is the time of flight for bullet?
(b) What is the muzzle velocity (velocity of bullet when it leaves the rifle)?

Answer 1

Answer:

The value is $u_x = 597 \ m/s$

Explanation:

From the question we are told that

    The distance of the target from the riffle is  $d = 40 \ m$

    The height at which the bullet hit the target is  $y = 2.2 \ cm = 0.022 \ m$

Considering the vertical motion  

Generally from kinematic equations we have that

        $y = u_y t + \frac{1}{2} * gt^2$

At the initial stage the velocity is  zero  i.e $u_y = 0 \ m/s$

=>     $0.022 = 0 * t + \frac{1}{2} * 9.8 t^2$

=>    $t = 0.067 \ s$

Generally the velocity of the bullet when it leaves the riffle is mathematically represented as

         $u_x = \frac{ d}{t}$

=>     $u_x = \frac{40 }{ 0.067 }$

=>     $u_x = 597 \ m/s$