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$_0$ ?

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


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


  • $\begingroup$ 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


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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.