# Why is the acceleration along $x$-axis during projectile motion equal to 0?

Why is the acceleration along x axis during projectile motion equal to $$0$$? And if it is equal to $$0$$ then the object shouldn't stop along x-axis after vertical velocity reaches $$0$$ along $$y$$-axis.

• And if it is equal to 0, then the object shouldn't stop along x-axis after vertical velocity reaches 0 along y-axis. Can you expand on this more? Why do you think this? Commented Jun 9, 2021 at 1:44
• write the equations and see why or why not
– Eli
Commented Jun 9, 2021 at 7:22

In a projectile motion, the only force acting on the particle in $$mg$$ downwards (along the negative y axis). Since the initial velocity of the particle is not along the direction of net force, we have to either resolve the initial velocity vector or the force vector into components. Breaking the initial velocity vector into components one along force(y -axis) and one perpendicular to the force(x -axis)(which is in this case parallel to the ground) to get that $$F_{net,x} = 0$$ (since the perpendicular component of a vector is a null vector, mathematically $$F cos90° = 0$$). So, $$a_x = 0$$.

And if it is equal to 0 then the object shouldnt stop along x-axis

After the vertical component of velocity reaches 0, the force decreases it (do not forget that at the surface of earth, force of gravity is independent of initial velocity and height. It is always acting on a body) thus making it to descend. After the particle touches the ground, the ground applies a contact force on the ball and deforms it a bit. This deformation leads to loss of kinetic energy of the particle(which we call as inelastic collision) thus stopping it after some time.

Hope it helps.

In projectile motion there is only one force, the force due to gravity. This force necessarily only acts downwards. By Newton's Second Law, $$F = ma$$, and since there is no force in the x-direction, acceleration in that direction is also zero.

The object doesn't "stop along the x-axis" because it started with some velocity in the x-direction. Since there is no acceleration, it keeps that velocity.

Yes it won't stop after landing. If the ground doesn't have any friction then it will continue to move on ground. But then the motion is along the straight line. The whole point of analysis of projectile motion is to understand the motion in 2-dimensions given the constraint that force is along negative y- direction. The motion on the ground after anding is just a 1-D motion. So in books, what happens after landing, is generally skipped as there prime concern is to show what happens when the body undergoing projectile motion in the air.

That's only while in the air when ignoring air resistance because there are no horizontal forces acting on it. When you hit the ground, things change. The ground is more than capable of applying forces in directions other than vertical. Nothing says x acceleration must remain zero forever and ever, especially when conditions change.

Similarly, a person free falling, ignoring terminal velocity, doesn't maintain 1g of acceleration even when they encounter the ground.

Why is the acceleration along x axis during projectile motion equal to 0?

Because gravity pulls you down.* Pick up an object, and let go (without giving it any push). It will fall straight down. It will not go sideways (assuming no wind of course). Gravity is a force that is directed downwards, and that is the direction along which it can accelerate -- downwards.

And if it is equal to 0 then the object shouldn't stop along x-axis after vertical velocity reaches 0 along y-axis.

No, because the $$x$$ velocity component -- whatever it is -- is unaffected by gravity.

*this simplification is only valid for projectile motions on a small scale. Since the earth is round, this simplification will break down if the projectile covers very long distances.