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when we calculate the time of flight for projectile with no drag we use second equation of motion i.e. $$S=V_i t + at^2/2$$ further, we put $V_i=V_i\sin\theta$ and $a=-g$ i.e.: $$S=V_i t\sin\theta - gt^2/2$$ further if we solve this equation it becomes $T= 2V_i \sin\theta/g$ from this formula it seems that this equation only calculate the time for vertical component of projectile (because $V_i\cos\theta$ is ignored). So it means that if a projectile is thrown at an angle this equation will not provide us an accurate information about the time spent by the projectile from projection point to the impact point! (is total time of flight is something different than what i thought?)

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    $\begingroup$ Because the flight will end when vertical displacement becomes zero. $\endgroup$ – William R. Ebenezer May 26 '19 at 7:51
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    $\begingroup$ If you defined the end of the trajectory to be when it has moved some horizonal distance then it would be different. It's just based on the definition of "end of flight". $\endgroup$ – BioPhysicist May 26 '19 at 10:04
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Because the horizontal dimension doesn't count. When a shell leaves the muzzle it starts to rise,then at the top of its parabola it starts to fall. It is this rising & falling which dictates the time of flight,along with air pressure (high air density retards progress of the missile).

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  • $\begingroup$ do u mean that the time spent in the air by the vertical component of motion is not added to the time of flight? $\endgroup$ – yaseen wazir May 26 '19 at 9:06
  • $\begingroup$ No,I mean the horizontal; dimension adds nothing to the time. $\endgroup$ – Michael Walsby May 26 '19 at 9:10
  • $\begingroup$ ohh...yeah yeah horzental...thanks $\endgroup$ – yaseen wazir May 26 '19 at 10:54
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Neglecting the influences of air, the only things that affect time in the air are the initial vertical component of velocity and the downward acceleration due to gravity. The initial horizontal component of velocity only affects the horizontal distance the shell goes before hitting the ground.

Hope this helps.

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Think of it this way. You can decompose the initial movement of the projectile into 2 components, one horizontal, the other vertical.

Now let's take an ideal case, where we ignore the resistance of the air, and assume the ground is a perfectly flat plane, not a globe like the earth.

In that case, the horizontal movement of the projectile will stay exactly the same as it was when it started, because no force affects it. The only movement that is affected by a force (gravity) is the vertical one. Hence, the projectile will hit the ground when gravity's interaction with the vertical component causes the height to return to zero. At the same time, the horizontal component, being unaffected by gravity, will dictate how far away from the starting point the projectile will land.

Air resistance will affect both of these, so the projectile will not travel as far, and will land sooner. The curvature of the earth will cause the projectile to land slightly later, as it has to travel further around the earth.

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  • $\begingroup$ i got this in khan acdmy: " if you fire a bullet horizontally and drop a bullet at the same time, they will hit the ground in the same time (if conditions like curvature , air resistance etc for the bullets are satisfied) $\endgroup$ – yaseen wazir May 26 '19 at 13:08
  • $\begingroup$ so it means that if the horizental bullet covers 1000000000km, then the other bullet which is just dropped from 1m above the ground, both will hit the ground at the same time...(very very counterintuitive!) $\endgroup$ – yaseen wazir May 26 '19 at 13:12
  • $\begingroup$ counter-intuitive, but correct! $\endgroup$ – hdhondt May 26 '19 at 23:30
  • $\begingroup$ got it...thanks! $\endgroup$ – yaseen wazir May 27 '19 at 2:18
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    $\begingroup$ Please accept my answer if you think it is the best one $\endgroup$ – hdhondt May 27 '19 at 4:14
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The horizontal velocity component does not count in case we neglect the Earth surface curvature and the air drag.

In case of at least the latter, vertical component of acceleration does not depend on horizontal component of velocity.

For very long shots by a super cannon on the Moon, horizontal velocity component would count as well due the Moon curvature.

Things get complicated by involving the air drag.

In this case, vertical acceleration is not independent on the horizontal velocity component ant more.

The faster the object flies horizontally, the stronger is uplifting vertical drag and the objects falls slower.

If we consider velocity components $$v_\mathrm{y}, v_\mathrm{x}=a \cdot v_\mathrm{y}$$ and if we consider the air drag force $$F=k\cdot v^2$$

then the vertical component of the air drag is :

$$\left| F_\mathrm{y}\right|=k \cdot v_\mathrm{y}^2 \cdot \sqrt{1+a^2}$$

( derivation in the hidden text )

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I think that you misunderstood it that it is the sum of the time spent by the projectile in air and on ground after its impact with the ground but it is defined as follows:-

The total time of flight is the time spent by the projectile in air.

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The time of flight is the time taken from projection to hitting the ground.

With no air resistance present and assuming level ground the initial velocity can be split into a vertical component and a horizontal component.

The vertical component of velocity undergoes an acceleration of $g$ downwards whereas the horizontal component of velocity is constant.

The distance traveled horizontally by the projectile depends on the horizontal speed of the projectile and the time that the projectile is in the air (time of flight).

The time that the projectile in in the air depends on the initial vertical speed of the projectile $v_{\rm vertical,initial}$, and the acceleration due to gravity, $g$ and can be found using the equation

$s = v_{\rm i} + \frac 12 a t^2 \Rightarrow 0 = v_{\rm vertical,initial}\,t-\frac 12 gt^2$ as the displacement $s$ in the vertical direction is zero.

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When determining how much time the projectile spends in the air it doesn't matter how much it moved sideways or not. The only thing that matters is how much time it took it to go up, slow down (vertically) due to gravity, and then go down (due to gravity) and touch the ground again. It's the same time irrelevant of if it went a 100 meters sideways, 1m, or not at all, IF and only IF of course if all these projectiles had the same vertical component of velocity and of course experienced the same gravity/were on the same planet.

You'd of course need more force to add the sideways motion in those other cases but that's a completely different topic.

Edit: Plus, I think you just misunderstood what "total time of flight" means. It's not "vertical time" plus "horizontal time", because it's a silly concept, they're the exact same thing, identical (distance is different, sure, time, no). It's just... time it spends in flight.

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