# Time of flight of projectile from launch to landing

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.

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

• You can find the proof in any good physics book. Also refer to Projectile_motion. – gimusi Nov 27 '18 at 22:04
• 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. – Edville Nov 27 '18 at 22:09
• Ah ok! I've added a hint to find the solution. – gimusi Nov 27 '18 at 22:14

• $$y(t)=h+v_0 \sin\theta \cdot t-\frac12 g t^2$$
then by the condition $$y(t)=0$$ find the time of landing $$t_{L}$$.
• $$d=x(t_{L})=v_0 \cos \theta \cdot t_{L}$$