A player kicks the ball. The ball attains a velocity of v m/s at a θº with respect to ground. If its trajectory begins and ends at the ground, how far from that player does it land? (ignore any drag or air resistance)

Question

A player kicks the ball. The ball attains a velocity of v m/s at a θº with respect to ground. If its trajectory begins and ends at the ground, how far from that player does it land?                                                                                                         (ignore any drag or air resistance)

idku · Answer

I know that:
$\large \color{royalblue}{{\rm V}_{ix}={\rm V}_{i}\cos\theta~~{\rm m/s}}$          and,
$\large \color{royalblue}{{\rm a}_{x}=0~~{\rm m/s^2}}$

(The velocity in the horizontal doesn't change,
because there is no acceleration, thus it turns out... )

$\large \color{royalblue}{{\rm V}_{fx}={\rm V}_{ix}+a_{x} t}$

$\large \color{royalblue}{\int {\rm V}_{fx}~{\rm d}t=\int{\rm V}_{ix}+a_{x} t~{\rm d}t}$

$\large \color{royalblue}{ {\rm D}_{x}={\rm V}_{ix}t+0.5~a_{x} t^2}$

$\large \color{royalblue}{ {\rm D}_{x}={\rm V}_{i}t\cos\theta}$

But when I given the angle of 36 degrees and initial velocity of 22.7 m/s, I am still missing the time.