Hello
The bullet is moving by uniformly accelerated motion.
The initial velocity is [tex]v_i=180~m/s[/tex], the final velocity is [tex]v_i=0~m/s[/tex], and the total time of the motion is [tex]\Delta t=0.02~s[/tex], so the acceleration is given by
[tex]a= \frac{v_f-v_i}{\Delta t} = -9000~m/s^2 [/tex]
where the negative sign means that is a deceleration.
Therefore we can calculate the total distance covered by the bullet in its motion using
[tex]S=v_i t + \frac{1}{2}at^2 = 180~m/s \cdot 0.02~s + \frac{1}{2}(-9000~m/s^2)(0.02~s)^2=1.8~m[/tex]
So, the bullet penetrates the sandbag 1.8 meters.
