Answer: 317.0 m/s
Explanation:
The motion of the bullet is a projectile motion, with:
- a uniform motion with constant speed v along the horizontal direction
- an accelerated motion with constant acceleration [tex]g=-9.81 m/s^2[/tex] toward the ground
We know that the starting height of the bullet is h=1.4 m. If we consider the vertical motion only, the initial velocity is zero, so we can write:
[tex]y(t)=h+\frac{1}{2}gt^2[/tex]
The bullet reach the ground when y(t)=0, so the time taken is
[tex]0=h+\frac{1}{2}gt^2\\t=\sqrt{-\frac{2h}{g}}=\sqrt{\frac{-2(1.4 m)}{-9.81 m/s^2}}=0.53 s[/tex]
During this time, the bullet travels d=168 m horizontally, so its horizontal speed (which is equal to the initial speed of the bullet) is given by
[tex]v=\frac{d}{t}=\frac{168 m}{0.53 s}=317.0 m/s[/tex]
Considering the parabolic moment, the time upto which the bullet was in air will be given by below equation:
Y = Y0 + V0y x t - 1/2 gt^2
Y0 = initial height
Y= Final height
0 = 1.4 + 0 - 0.5 x 9.8 m/s^2 x t^2
t= 0.533 seconds
Speed of bullet = distance traveled /time taken
= 168 m/0.533 seconds
= 315 m/s
