The horizontal distance traveled by the ball (range of motion) is given by the following equation:
[tex]x= v_{0x} t[/tex]
In which [tex]x[/tex] is the range of motion, [tex] v_{0x} [/tex] is the horizontal component of the initial velocity, and [tex]t[/tex] is the time of motion.
First, lets calculate the horizontal component of the initial velocity:
[tex] v_{0x}= v_{0}cos( \alpha)=13.9cos(35)=11.39[/tex]
Now, we calculate the time of motion from the equation that described the motion of the ball in the vertical axis:
[tex]y= \frac{1}{2}at^2+ v_{0y}t [/tex]
In which [tex]y[/tex] is the position of the ball vertically, [tex] v_{0y} [/tex] is the vertical component of the initial velocity, [tex]a[/tex] is the acceleration in the vertical axis (which is gravity), and [tex]t[/tex] is time of motion.
We want to find the time when the ball lands, hence, when [tex]y=0[/tex]; so the equation becomes:
[tex]0= \frac{1}{2}at^2+ v_{0y}t=\frac{1}{2}at^2+ v_{0}sin( \alpha )t=\frac{1}{2}(-9.8)t^2+ 13.9sin(35 )t[/tex]
We rewrite it a bit more:
[tex]-4.9t^2+7.97t=0[/tex]
This is a quadratic equation, so we use the quadratic equation formula to solve for time (we'll get two answers):
[tex]t= -\frac{1}{9.8} [{-7.97}+-\sqrt{(7.97)^2}][/tex]
Clearly, one of the answers is [tex]t=0[/tex], this is before you kick the ball (it is on the ground), we want the nonzero answer (when it lands) so:
[tex]t= -\frac{1}{9.8} ({-7.97}-7.97})=1.63[/tex]
Now, we plug-in the time value to the equation of the motion's range:
[tex]x= v_{0x} t=(11.39)(1.63)=18.57[/tex]
The ball will travel 18.57 meters.
