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,
![h=u_yt+(1)/(2)gt^2](https://img.qammunity.org/2023/formulas/physics/college/vu2z7nhjuzfmsb3z79unwjox9iujv3cmse.png)
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,
![\begin{gathered} 22=0+(1)/(2)*9.8* t^2 \\ t^2=(22*2)/(9.8) \\ t^2=4.49 \\ t=2.12\text{ s} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/y23698b3aofltb8qbid0j8957nhkbuj8bx.png)
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,
![R=u_xt+(1)/(2)at^2](https://img.qammunity.org/2023/formulas/physics/college/tzo69hzb4a5r9jq52aiipbti2o2gsqmwot.png)
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,
![\begin{gathered} R=524*2.12+0 \\ R=1110.9\text{ m} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/grqrjn2eks5iiyhwtnzenax76uc1q5qhzg.png)
Thus, the horizontal range of the bullet is 1110.9 meters.
Hence, the bullet hit the ground at 1110.9 meters.