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The formula below is the time of flight ( time of whole journey from launch(0,0) to landing (×,y) ) of a projectile whose initial vertical position is above the point of impact.

enter image description here

I am trying to understand how the right side of the equation is derived. For instance, how do I come up with 2gy$_0$ ?

$\frac{d}{v.cos(\theta)}$ = $\frac{v.sin(\theta)+\sqrt{\left(v.sin(\theta)\right)^2 +2gy_0}}{g}$

Where

g = gravitational acceleration

y$_0$ = initial vertical position (h)

d = entire horizontal distance or range of the flight from launch to landing

v = velocity

$\theta$ = initial launch angle

Thanks

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  • $\begingroup$ You can find the proof in any good physics book. Also refer to Projectile_motion. $\endgroup$ – gimusi Nov 27 '18 at 22:04
  • $\begingroup$ This was actually from wikipedia. It didn't show how the right hand side is derived, and I am unable to find an online resource for this particular derivation. $\endgroup$ – Edville Nov 27 '18 at 22:09
  • $\begingroup$ Ah ok! I've added a hint to find the solution. $\endgroup$ – gimusi Nov 27 '18 at 22:14
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HINT

Let consider at first the equation of motion in vertical direction that is

  • $y(t)=h+v_0 \sin\theta \cdot t-\frac12 g t^2$

then by the condition $y(t)=0$ find the time of landing $t_{L}$.

Finally use that to find x of landing by

  • $d=x(t_{L})=v_0 \cos \theta \cdot t_{L}$
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