# Projectile Motion and acceleration of particle

## Homework Statement

If a particle moves in X-Y plane with acceleration non zero in X and Y , the particle will not move in a parabolic path
True or False ?

## Homework Equations

X = UxT + 1/2(Ax)T^2
Y= UyT + 1/2 (Ay)T^2

## The Attempt at a Solution

The equation of trajectory that i came up with involved Y^2 , X^1/2 and X
I am not able to draw conclusions with it
So i guessed it to be True as for a parabolic path you need constant acc in only one axis and zero acc in other axis
But the book says false

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berkeman
Mentor
Welcome to the PF.

Think about how a projectile traces out a parabolic shape when you throw it up and out at a 45 degree angle. What forces and accelerations are acting on it in that situation (after it leaves your hand)?

Welcome to the PF.

Think about how a projectile traces out a parabolic shape when you throw it up and out at a 45 degree angle. What forces and accelerations are acting on it in that situation (after it leaves your hand)?
Welcome to the PF.

Think about how a projectile traces out a parabolic shape when you throw it up and out at a 45 degree angle. What forces and accelerations are acting on it in that situation (after it leaves your hand)?
Just the gravity pull and air friction

berkeman
Mentor
Just the gravity pull and air friction
Good. So neglecting air friction (which will alter the path away from a pure parabola), in how many dimensions is the projectile accelerating while it traces out the parabola?

Good. So neglecting air friction (which will alter the path away from a pure parabola), in how many dimensions is the projectile accelerating while it traces out the parabola?
2 dimensional motion with downward acceleration (g)

berkeman
Mentor
@AkshayM -- You marked this thread as solved -- does that mean you understand now?

@AkshayM -- You marked this thread as solved -- does that mean you understand now?
Sorry no bymistakely i did it

berkeman
Mentor
If a particle moves in X-Y plane with acceleration non zero in X and Y , the particle will not move in a parabolic path
True or False ?
But the book says false
So the book is saying that the path is parabolic even when there is acceleration in 2 dimensions?

So the book is saying that the path is parabolic even when there is acceleration in 2 dimensions?
Yes it says parabolic path
Case of a misprint ?

berkeman
Mentor
Maybe. Can you ask the professor or a TA about it? Do you know any other students in your class who are also working on the problem?

haruspex
Homework Helper
Gold Member

## Homework Statement

If a particle moves in X-Y plane with acceleration non zero in X and Y , the particle will not move in a parabolic path
True or False ?

## Homework Equations

X = UxT + 1/2(Ax)T^2
Y= UyT + 1/2 (Ay)T^2

## The Attempt at a Solution

The equation of trajectory that i came up with involved Y^2 , X^1/2 and X
I am not able to draw conclusions with it
So i guessed it to be True as for a parabolic path you need constant acc in only one axis and zero acc in other axis
But the book says false
First, let us be clear that the question is asking whether a nonzero acceleration in both coordinates guarantees that it will not be a parabola. My point is nothing has been said about constancy of acceleration, so certainly it might not be a parabola.

So now suppose there is constant acceleration in each coordinate. What if you were to use different axes? Might there be a direction in which there is no acceleration? Will there always be such a direction?

Ray Vickson
To re-express the answer of haruspex in post #11: if you take the path to be $x = v_x t + \frac{1}{2} a_x t^2$ and $y = v_y t + \frac{1}{2} a_y t^2,$ (with constants $v_x, v_y, a_x,a_y$) can you find a constant angle $\theta$ and a rotated coordinate system
$$\begin{array}{rcl} X&=& x \cos(\theta) - y \sin (\theta)\\ Y&=& x \sin (\theta) + y \cos (\theta) \end{array}$$ in which the new path has the form $X = U_x t, \; Y = U_y t + \frac{1}{2} W_y t^2,$ (with constant $U_x, U_y, W_y$) or is that not possible? If it is possible, that would give you a "rotated" parabola.