<s>[INST] How would you find the kinematic range using the position function? Knowing that the range is: 
$$ R = \frac{v^2\sin2\theta}g $$
Taking the integral of the velocity function we have:
$$ R(T) = (V_i \cos\theta T + x_i)X +\left(-\frac{1}2gT^2+V_i\sin\theta T+y_i\right)Y$$
So, I know it hits the ground at:
$$ \text{time in flight} = \frac{V_i\sin\theta \pm \sqrt{ (V_i\sin\theta )^2 +2gy_i } }g $$
Knowing the time in flight, how would we derive the range?
Would I simply plug it back into the position function on the $x$-axis? (Assuming the object hits the ground when $y = 0$)
Edit:
After messing with it, I found out what the book wanted me to do:
Assuming $y_i = 0:$
$ timeInFlight = \frac{ V_isin\theta + \sqrt{ (V_isin\theta)^2 } }g = \frac{ 2V_isin\theta }g $
$ R_x(T)=V_icos\theta T+x_i $
$R_x(time in flight)  = \frac{2V_i^2sin\theta cos\theta}g + x_i = \frac{V_i^2sin2\theta}g + x_i$
 [/INST] \begin{align*}
  x &= ut\cos \theta \\
  y &= ut\sin \theta -\frac{gt^{2}}{2} \\
    &= x\tan \theta -\frac{gx^{2}}{2u^{2}\cos^{2} \theta} \\
    &= x\tan \theta \left( 1-\frac{gx}{2u^{2} \sin \theta \cos \theta} \right) \\
    &= x\tan \theta \left( 1-\frac{gx}{u^{2} \sin 2\theta} \right) \\
  R &= \frac{u^{2} \sin 2\theta}{g}
\end{align*}
</s>