# 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 $y_i = 0:$

$timeInFlight = \frac{ V_isin\theta + \sqrt{ (V_isin\theta)^2 } }g = \frac{ 2V_isin\theta }g$

$R_x(T)=V_icos\theta T+x_i$

$R_x(time in flight) = \frac{2V_i^2sin\theta cos\theta}g + x_i = \frac{V_i^2sin2\theta}g + x_i$

Hint your formula for range is wrong its $$\frac{v^2\sin(2\theta)}{g}$$ and for knowing range using time of flight we can use $s=v_x.t+\frac{1}{2}gt^2$ where $v_x$ is velociyy in x direction . s is the displacement. Now we know time of flight is $2v\sin(\theta)/g$ just plug in and get range ie s
• Then positions are given by $ucos\theta.t, usin\theta.t-gt^2/2$ – Archis Welankar Feb 17 '16 at 6:23
• Also, the TIF = $\frac{2Vsin(\theta)}g$ You wrote it as v^2 – Adam Reed Feb 17 '16 at 6:34