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Assuming some random object (spherical cow/brick) reaches a speed v in free fall:

if you pick this same object and you accelerate it horizontally up to the same speed v in the same type of atmosphere, pointing the same side of it in the direction of movement,

would this object be able to maintain flight?

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  • $\begingroup$ I assume you mean that $v$ is the object's terminal speed. Are you maintaining a constant force on the object? $\endgroup$ – PM 2Ring May 23 at 8:26
  • $\begingroup$ @PM2Ring: yes, that is. $\endgroup$ – Quora Feans May 23 at 8:38
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When an object moves through a fluid, in this case the air, the fluid exerts a force on it which is call lift. The lift force is perpendicular to the direction of motion, but it can point in any direction contained on the plane perpendicular to the direction of motion. The particular direction depends on the shape of the object. For example the wings of planes are made in a way that the lift force acts upwards (aggainst gravity) allowing them to fly, but racing cars are designed such that the lift force acts downwards so that the car is more attached to the ground. So the answer is that it depends on the shape.

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  • $\begingroup$ I was thinking that if a spherical object is already at terminal speed (horizontally in this case) it would need to move faster than that to have a vertical speed component. So, I thought it was already at a stable position and would remain there. $\endgroup$ – Quora Feans May 23 at 9:20
  • $\begingroup$ But the concept of termial speed can just be done in free fall, since it is by definition the maximum velocity that you achieve in free fall (when the air resistance force equals the gravity force). In the horizontal case there is no terminal speed in this sense. However we can talk about the speed at which the lift force will be greater than the gravity force and the object will start to fly if you want, and call it whatever. Then you are right if the object does not exceed this "terminal velocity" it won't fly. $\endgroup$ – vin92 May 23 at 9:54
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Your object would have 2 completely separate motions:

  1. Its horizontal movement. In the absence of air friction it will keep that movement unchanged.
  2. Its vertical movement. Even if it was stationary (vertically) when you launched it, as soon as you release it it will start falling with an acceleration of $9.8m/s^2$. So after $1s$ it will be travelling downwards at $9.8 m/s$ and it will keep accelerating.

As a result the object will follow a parabola in which it moves at the same speed horizontally until it finally hits the ground. So no, it will not stay up and "maintain flight".

Newton was the first to illustrate this will the thought experiment of a cannonball being launched at faster and faster speeds, until it finally goes so fast that it "falls" right around the earth. Which is how satellites stay in orbit: they are always accelerating to the earth at $9.8 ms/^2$ (actually a little less because they are further from the earth than we are) but they are moving at $7.8 km/s$ so they continuously miss the earth.

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  • $\begingroup$ The object isn't following a ballistic trajectory. A constant force is being applied to it as it travels (in addition to the gravitational force). $\endgroup$ – PM 2Ring May 23 at 16:05
  • $\begingroup$ @PM2Ring The question states "you accelerate it horizontally up to the same speed". I assume it then keeps that speed without further acceleration. $\endgroup$ – hdhondt May 24 at 0:09
  • $\begingroup$ I asked the OP: "Are you maintaining a constant force on the object?" and they replied "yes". $\endgroup$ – PM 2Ring May 24 at 5:09
  • $\begingroup$ @PM2Ring The original question did not say that, and I suspect the OP is confused between force and speed. If the same force is maintained, then the speed will increase forever. Once it reaches escape velocity the object will indeed "maintain flight". Accelerating "horizontally up to the same speed v" implies the force disappears - or is reduced to the force necessary to overcome air friction and maintain that speed. $\endgroup$ – hdhondt May 24 at 5:50

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