Question

why is the no acceleration in x componet during projectile

Answer 1

if a picture would help, lemme know--the answer below is a copy of one i made a while ago.

With projectile motion, we often choose the y direction to be vertical--this is the direction in which gravity acts! We often choose the x direction as horizontal. The kind of math we use in Physics (vectors) allows us to split 2d motion up into two SEPARATE pieces which are perpendicular to each other and NOT AFFECT EACH OTHER AT ALL. because the x (left-right) and y (up-down) directions are perpendicular, we can look at them SEPARATELY. SO the influence of gravity, which works in the y direction, does NOT effect the x direction.

There is no acceleration due to gravity in the x direction. What else could case an acceleration in the x direction? Well, air resistance could, but we often IGNORE air resistance. SO--there is NO ACCELERATION IN THE X DIRECTION!

Answer 2

In the projectile motion you can describe it by considering two motions along the two axis: one, along x, is a uniform motion (constant velocity, zero acceleration) the other, along y, its a accelerated motion, increasing velocity with g acceleration.

If you use the same equation \[x = x_0 + V_0 \times t + \frac{ 1 }{ 2 } \times g \times t²\] in this two different situations for the same projectile motion you are describing accurately this type of motion.

In the particular case of shooting a ball towards your front with \[V_0\] velocity in the beginning of your reference axis and falls...

(see the image attached)

you get along x: \[x = V_0 \times t \] where, \[x_0 = 0 m\], and \[V_0\] for the uniform motion (without acceleration)

Along y: \[y = \frac{ 1 }{ 2 } \times g \times t²\] with where, \[y_0 = 0 m\], and \[V_0 = 0\ m/s\], the initial velocity, in this particular case, its along x only. If you have a velocity along y, then you should put it also on this last equation;

So, using this two equations you can describe this no drag projectile motion: \[y = \frac{ 1 }{ 2 } \times g \times t²\] and \[x = V_0 \times t \].

Answer 3

that a simple one because there is no force acting in the x axis there is no acceleration by newtons law