Given data:
* The height of the stone is h = 50 m.
* The initial horizontal speed of the stone is v = 4 m/s.
Solution:
By the kinematics equation along the vertical direction, the time taken by the stone to reach the ground is,
![h=ut+(1)/(2)gt^2](https://img.qammunity.org/2023/formulas/physics/college/yhpvuknn8t5u6t9cq21om3ae6aded6uwqf.png)
where u is the initial vertical speed of the stone,
The initial vertical velocity of the stone is zero, thus,
![h=(1)/(2)gt^2](https://img.qammunity.org/2023/formulas/physics/college/kfh0snue2m894wjq67wbih9op261o60363.png)
where g is the acceleration due to gravity,
Substituting the known values,
![\begin{gathered} 50=(1)/(2)*9.8* t^2 \\ t^2=(2*50)/(9.8) \\ t^2=10.2 \\ t=\pm3.19\text{ s} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/232pjbev9ilw3r30d1rlyzgs8ddf4prp0p.png)
Neglecting the negative value of time,
Thus, the time taken by the stone to reach the ground is 3.19 s.
Hence, option D is the correct answer.