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 $y = 0$)


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

  • $\begingroup$ R(T) is the position of the object. Regular R= is the range equation. It was taken directly out of my book. (Also, I meant sin(2x) not sin ^2(x). Just fixed it. $\endgroup$ – Adam Reed Feb 17 '16 at 6:14
  • $\begingroup$ Then positions are given by $ucos\theta.t, usin\theta.t-gt^2/2$ $\endgroup$ – Archis Welankar Feb 17 '16 at 6:23
  • 1
    $\begingroup$ Also, the TIF = $ \frac{2Vsin(\theta)}g $ You wrote it as v^2 $\endgroup$ – Adam Reed Feb 17 '16 at 6:34

\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*}


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