Answer: 1.12 m
Explanation:
This situation is related to parabolic motion, hence we can use the following equations:
[tex]y=y_{o}+V_{o}sin \theta t-\frac{g}{2}t^{2}[/tex] (1)
[tex]x=V_{o} cos \theta t[/tex] (2)
Where:
[tex]y=0 m[/tex] is the ball final height (when it hits the ground)
[tex]y_{o}=1.1 m[/tex] is the ball initial height
[tex]V_{o}=2.2 m/s[/tex] is the initial velocity
[tex]\theta=30\°[/tex] is the angle at which the ball was launched
[tex]t[/tex] is the time
[tex]g=9.8 m/s^{2}[/tex] is the acceleration due gravity
[tex]x[/tex] is the horizontal distance the ball travels
Rewriting (1) with the given values:
[tex]0 m=1.1 m+(2.2 m/s)(cos 30\°)t-\frac{9.8 m/s^{2}}{2}t^{2}[/tex] (3)
Multiplying all the eqquation by -1 and rearranging:
[tex]4.9 m/s^{2} t^{2}-1.1 m/s t-1.1 m=0[/tex] (4)
So, since we have a quadratic equation here (in the form of[tex]0=at^{2}+bt+c[/tex], we will use the quadratic formula to find [tex]t[/tex]:
[tex]t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}[/tex] (5)
Where [tex]a=4.9[/tex], [tex]b=-1.1[/tex], [tex]c=-1.1[/tex]
Substituting the known values and choosing the positive result of the equation, we have:
[tex]t=\frac{-(-1.1)\pm\sqrt{(-1.1)^{2}-4(4.9)(-1.1)}}{2(4.9)}[/tex]
[tex]t=0.59 s[/tex] (6)
Now, substituting (6) in (2):
[tex]x=(2.2 m/s)(cos 30\°)(0.59 s)[/tex] (7)
[tex]x=1.12 m[/tex] (8) This is the horizontal distance at which the ball hits the ground.
