Given data:
* The initial speed of the projectile is u = 20 m/s.
* The angle of the projectile with the horizontal is,
[tex]\theta=25^{\circ}_{}[/tex]
Solution:
The initial horizontal component of the speed is,
[tex]u_x=u\cos (\theta)[/tex]
Substituting the known values,
[tex]\begin{gathered} u_x=20\cos (25^{\circ}) \\ u_x=20\times0.9063 \\ u_x=18.126\text{ m/s} \\ u_x\approx18.13\text{ m/s} \end{gathered}[/tex]
As no force acting on the projectile along the horizontal direction.
According to Newton's second law, the acceleration along the horizontal direction is,
[tex]\begin{gathered} F=ma \\ a=\frac{F}{m} \\ a=0ms^{-2} \end{gathered}[/tex]
Thus, no change in the velocity takes place along the horizontal direction.
Hence, the final horizontal component of velocity before striking the ground is 18.13 m/s.
