# Does horizontal acceleration affect gravity?

If we apply 1G horizontally in some object, will this constant force equal to G affect the time of falling? If the force does not affect gravity, why gravity is prioritized over this force if both are equal?

Edited: For the ones who didn't understand what i mean well, i mean why we can't switch them and say gravity is the horizontal force, and the force we applied is gravity? why the trajectory does not change as shown in this image?:

• Gravity is not prioritised over this horizontal force. Rather, they apply simultaneously. You still land after the same amount of time. But you land much farther to the side. That is the combined effect. Nov 29, 2022 at 13:00
• Do you mean apply a horizontal force of magnitude $mg$? You would simply have an object in constant acceleration down and in that direction, $\vec{a} = (g, -g)$ in a standard $x$-$y$ coordinate system. You could determine the time of flight using $H = \frac{1}{2} g t^2$ and the final horizontal distance traveled by $x = \frac{1}{2} g t^2$. Nov 29, 2022 at 13:26
• I've made an update. Nov 29, 2022 at 14:10
• I have no idea what your pictures mean. Did you understand what I said above? A force of "1G" doesn't have any meaning. Nov 29, 2022 at 14:30
• @BenH why the trajectory of the projectile does not happen as in the second image? Nov 29, 2022 at 16:43

According to vector projection rules,

$$a_{1}=\left\|\mathbf {a} \right\|\cos \theta \tag 1,$$

when $$\vec a_1 \perp \vec a$$ , as so it is in your case, angle between force (or acceleration) vectors is $$\pi/2$$,- then (1) gives :

$$a_{1}=\left\|\mathbf {a} \right\|\cos \frac {\pi}{2} = 0 \tag 2,$$

And hence, perpendicular forces do not affect each other in any direct way. So, the answer would be that body would still fall down with same $$1g$$ acceleration, despite the fact that you'll push it horizontally with $$1000g$$ at the same time.

If the force does not affect gravity, why gravity is prioritized over this force[...]?

None of forces are "prioritized". Simply each force has "line of action". When no other forces has vector projection $$\gt 0$$ along that line of action (in this case downwards),- then only that single force (in this case- gravity) remains. Otherwise, one needs to compute a net force in some direction to get the net acceleration in that direction.

• What will the trajectory be like? Nov 30, 2022 at 12:04
• If this applied force acts over all duration until object will reach ground (and I assume it is the case because you wrote in a question that this applied force is "constant", i.e. it acts not just at initial time point for setting object kinetic energy),- then trajectory will be just a straight line directed from upper-right to bottom-left. This is due to net force vector addition. So, neither case as you painted. Parabolic trajectories are only in case when object is pushed just in one direction, but it had initial kinetic energy in other directions (like when you kick a ball from a roof). Nov 30, 2022 at 12:51

Let's assume we have an object of mass $$m$$ with two forces acting: gravity vertically down with magnitude $$F_g = mg$$, and a constant applied horizontal force (say, to the left, as in the OP's picture) with the same magnitude, $$F_{\rm app} = mg$$. Just looking at the free-body diagram, we see that the net force, and thus the acceleration, must point down and to the left. And because the forces are both constant, the acceleration will also be constant.

Let's choose an $$x$$-$$y$$ coordinate system with $$x$$ pointing left and $$y$$ pointing down. Then Newton's Second Law gives: \begin{align} \vec{F}_{\rm net} &= m \vec{a}\\ \vec{F}_g + \vec{F}_{\rm app} &= m \vec{a}\\ x: \quad F_{g,x} + F_{{\rm app},x} &= m a_x \\ 0 + F_{\rm app} &= m a_x \\ m g &= m a_x \quad \rightarrow \quad a_x = g\\ y: \quad F_{g,y} + F_{app,y} &= m a_y \\ F_g + 0 &= m a_y \\ m g &= m a_y \quad \rightarrow \quad a_y = g \end{align} So the constant acceleration vector is $$\vec{a} = (a_x, a_y) = (g, g)$$.

Now if acceleration is constant, the equations for motion with constant acceleration can be used. In the $$x$$ direction we have: \begin{align} x(t) &= x_0 + v_{0x} t + \frac{1}{2} a_x t^2 \\ &= 0 + 0 + \frac{1}{2} g t^2 \end{align} and \begin{align} y(t) &= y_0 + v_{0y} t + \frac{1}{2} a_y t^2 \\ &= 0 + 0 + \frac{1}{2} g t^2 \end{align} where I assumed that the particle starts at the origin with zero initial velocity, $$\vec{v}_0 = (v_{0x}, v_{0y})= (0,0)$$.

The trajectory is the path we see it take through space, i.e., the function $$y(x)$$. So we must eliminate time from the above two equations: $$x(t) = \frac{1}{2} g t^2 \quad \rightarrow \quad y(t) = \frac{1}{2} g t^2 = x(t)$$ or, $$y(x) = x$$ The trajectory is a straight line down and to the left with slope 1.

Note that if you wanted to find the time for the object to hit the ground, you would just use the $$y$$ equation: $$y(t) = \frac{1}{2} g t^2 \quad \rightarrow \quad H = \frac{1}{2} g t_{\rm hit}^2 \quad \rightarrow \quad t_{\rm hit} = \sqrt{\frac{2H}{g}}$$ which is exactly what you would find for an object dropped from the same height. So, no, the horizontal force does not change the time of fall.

In newtonian mechanics atleast we take acceleration to be a vector, and in english we can usually say that gravity acts vertically, so when you have a force that is acting horizontally that is 90 degrees relative to the vertical this leaves no component of the force to decelerate the falling gravitational effect.
I hope that helps.

• Will the trajectory change? Nov 30, 2022 at 12:03
• If the object starts at rest then we usually say it doesn't unless there is something that we haven't factored for. Is there something that has inspired this question? i.e. something you have observed? Dec 6, 2022 at 8:07

The vertical component of net acceleration which is equal to $$g$$ directed vertically downward is prioritized because this component can only decide the time of flight ,as soon as the distance of projectile from the ground becomes zero then we consider as it has stopped going forward, which mainly is ruled by this component that when is the projectile going to complete its trajectory.

(Furthermore after the collision bouncing may happen depending on the coefficient of restitution between projectile and the ground, but this is not the case we are talking about, here we are considering only a single parabolic trajectory motion)

Else the horizontal component of net acceleration will only decide the range of that projectile.

We have Newton's second law F = ma which is a vector equation. If we look at the horizontal and vertical components, gravity acts vertically and your 1g force acts horizontally. So the vertical acceleration is determined by gravity. The 1g force has no vertical component so does not affect the vertical motion.

In your terms, gravity is prioritized in determining the time of fall because it's the only force in the vertical direction.