Answer: 2 m
Explanation:
We can solve this problem related to parabolic motion with the following equations:
[tex]y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2}[/tex] (1)
[tex]x=V_{ox}t[/tex] (2)
Where:
[tex]y=0 m[/tex] is the ball's final jeight
[tex]y_{o}=1.25 m[/tex] is the ball's initial height
[tex]V_{oy}=0 m/s[/tex] is the ball's initial vertical velocity, since the ball is horizontally projected
[tex]t[/tex] is the time
[tex]g=9.8 m/s^{2}[/tex] is the acceleration due gravity
[tex]x[/tex] is the ball's horizontal range
[tex]V_{ox}=4m/s[/tex] is the ball's initial horizontal velocity
Isolating [tex]t[/tex] from (1):
[tex]t=\sqrt{\frac{2 y_{o}}{g}}[/tex] (3)
[tex]t=\sqrt{\frac{2 (1.25 m)}{9.8 m/s^{2}}}[/tex] (4)
[tex]t=0.5 s[/tex] (5)
Substituting (5) in (2):
[tex]x=(4 m/s)(0.5 s)[/tex] (6)
Finally we find how far horizontally the ball lands:
[tex]x=2 m[/tex] (7)
