Given data:
* The height of the bullet is 22 m.
* The speed of bullet in the horizontal direction is 524 m/s.
Solution:
By the kinematics equation, the time taken by the bullet to reach the ground is,
[tex]h=u_yt+\frac{1}{2}gt^2[/tex]where u_y is the vertical velocity component, t is the time taken to reach the ground, g is the acceleration due to gravity, and h is the height of the bullet,
Substituting the known values,
[tex]\begin{gathered} 22=0+\frac{1}{2}\times9.8\times t^2 \\ t^2=\frac{22\times2}{9.8} \\ t^2=4.49 \\ t=2.12\text{ s} \end{gathered}[/tex]Thus, the time taken by the bullet to reach the ground is 2.12 seconds.
By the kinematics equation for the horizontal motion, the horizontal range of the bullet is,
[tex]R=u_xt+\frac{1}{2}at^2[/tex]where u_x is the horizontal component of the velocity, a is the acceleration along the horizontal direction, t is the time taken to reach the ground and R is the horizontal range,
Substituting the known values,
[tex]\begin{gathered} R=524\times2.12+0 \\ R=1110.9\text{ m} \end{gathered}[/tex]Thus, the horizontal range of the bullet is 1110.9 meters.
Hence, the bullet hit the ground at 1110.9 meters.