Projectile Motion and acceleration of particle

In summary: 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 (\the
  • #1
AkshayM
5
0

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
Please help !
 
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  • #2
Welcome to the PF. :smile:

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)?
 
  • #3
berkeman said:
Welcome to the PF. :smile:

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)?
berkeman said:
Welcome to the PF. :smile:

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
 
  • #4
AkshayM said:
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? :smile:
 
  • #5
berkeman said:
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? :smile:
2 dimensional motion with downward acceleration (g)
 
  • #6
@AkshayM -- You marked this thread as solved -- does that mean you understand now?
 
  • #7
berkeman said:
@AkshayM -- You marked this thread as solved -- does that mean you understand now?
Sorry no bymistakely i did it
 
  • #8
AkshayM said:
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 ?
AkshayM said:
But the book says false
So the book is saying that the path is parabolic even when there is acceleration in 2 dimensions?
 
  • #9
berkeman said:
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 ?
 
  • #10
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?
 
  • #11
AkshayM said:

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
Please help !
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?
 
  • #12
AkshayM said:
Yes it says parabolic path
Case of a misprint ?
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.
 
Last edited:

1. What is projectile motion and how does it differ from regular motion?

Projectile motion is the motion of an object that is launched into the air and then moves along a curved path under the influence of gravity. It differs from regular motion because it involves both horizontal and vertical components of motion, whereas regular motion only involves one direction.

2. What is the formula for calculating the acceleration of a particle in projectile motion?

The formula for calculating the acceleration of a particle in projectile motion is a = -g*sin(theta), where a is the acceleration, g is the acceleration due to gravity (9.8 m/s^2), and theta is the angle of launch.

3. How does the angle of launch affect the acceleration of a particle in projectile motion?

The angle of launch directly affects the acceleration of a particle in projectile motion. The higher the angle of launch, the greater the vertical component of the particle's initial velocity, resulting in a larger acceleration due to gravity. The lower the angle of launch, the smaller the vertical component and therefore a smaller acceleration due to gravity.

4. What is the maximum height reached by a projectile and how is it calculated?

The maximum height reached by a projectile is the highest point on its curved path. It can be calculated using the formula h = (v^2 * sin^2(theta)) / 2g, where h is the maximum height, v is the initial velocity, theta is the angle of launch, and g is the acceleration due to gravity.

5. Can the acceleration of a particle in projectile motion ever be zero?

Yes, the acceleration of a particle in projectile motion can be zero at the highest point of its trajectory, where it momentarily stops moving vertically and begins to accelerate downward due to gravity. At this point, the vertical component of the particle's velocity is zero, resulting in zero acceleration.

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