Titled reference frame, N2L with position and velocity

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SUMMARY

The discussion focuses on analyzing the motion of a ball thrown up an inclined plane using Newton's Second Law (N2L). The ball's position as a function of time is derived, leading to the formula for the range, R = 2vo^2(sin(theta)cos(theta+fi))/gcos^2(fi), and the maximum range, Rmax = vo^2/g(1+sin(fi)). Participants clarify the components of velocity and acceleration along and perpendicular to the incline, emphasizing the importance of correctly applying kinematic equations to find the range of the projectile.

PREREQUISITES
  • Understanding of Newton's Second Law (N2L)
  • Familiarity with kinematic equations
  • Knowledge of projectile motion concepts
  • Basic trigonometry, particularly sine and cosine functions
NEXT STEPS
  • Study the derivation of projectile motion equations on inclined planes
  • Learn about the components of forces acting on inclined surfaces
  • Explore advanced kinematics involving angular motion
  • Investigate the effects of different launch angles on projectile range
USEFUL FOR

Physics students, educators, and anyone interested in understanding projectile motion and its applications in real-world scenarios, particularly in the context of inclined planes.

Oblio
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A ball is thrown with initial speed vo up an inclined plane. The plane is inclined at an angle(fi) above the horizontal, and the ball's initial vecity is at an angle (theta) above the plane. Choose the axes with x measured up the slope, y normal to the slope and z across it. Write down Newton's second law using these axes and find the ball's position as a function of time. Show that the ball lands a distance

R=2vo^2(sin(theta)cos(theta+fi))/gcos^2(fi)

from its launch point. Show that for given vo adn (fi), the maximum possible range up the inclined plane is

Rmax=vo^2/g(1+sin(fi))

[Don't know if I'm spelling fi right, I mean this symbol : \phi



How do I approach this? I think the question is written poorly, they mean that the ball is thrown in the air and it lands ON the plane.
Without knowing the velocity how do I find the parabolic motion the ball will make?
 
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I know that N2L for the ball can't just be ma=-mg (or a= -g) since it's still traveling up...
 
Oblio said:
A ball is thrown with initial speed vo up an inclined plane. The plane is inclined at an angle(fi) above the horizontal, and the ball's initial vecity is at an angle (theta) above the plane. Choose the axes with x measured up the slope, y normal to the slope and z across it. Write down Newton's second law using these axes and find the ball's position as a function of time. Show that the ball lands a distance

R=2vo^2(sin(theta)cos(theta+fi))/gcos^2(fi)

from its launch point. Show that for given vo adn (fi), the maximum possible range up the inclined plane is

Rmax=vo^2/g(1+sin(fi))

[Don't know if I'm spelling fi right, I mean this symbol : \phi



How do I approach this? I think the question is written poorly, they mean that the ball is thrown in the air and it lands ON the plane.
Without knowing the velocity how do I find the parabolic motion the ball will make?
?? You know the balls inital velocity! It is v0. Further, since you are also told that the ball is initially thrown at angle \theta above the plane you know that the balls initial x component of velocity is v_0 cos(\theta) and the y component is v_0 sin(\theta). You should be able to show that the balls position at time t is x= v_0 cos(\theta) t and y= -g/2 t^2+ v_0 sin(/theta)t (taking (0,0) as initial position). You can solve the first equation for t: t= x/(v_0 cos(\theta) and put that into the equation for y: y= -g/(2v_0 cos(\theta)x^2+ (sin(\theta)/cos(\theta))x to get the equation of the parabola. Now it is a question of determing where that parabola intersects the line giving the inclined plane. That is, of course, y= tan(\phi)x= (sin(\phi)/cos(\phi))x.
 
\phi is spelt 'Phi'
 
HallsofIvy said:
y= -g/2 t^2+ v_0 sin(/theta)t (taking (0,0) as initial position). You can solve the first equation for t: t= x/(v_0 cos(\theta) and put that into the equation for y: y= -g/(2v_0 cos(\theta)x^2+ (sin(\theta)/cos(\theta))x to get the equation of the parabola. , y= tan(\phi)x= (sin(\phi)/cos(\phi))x.

Can you explain to me how you got y= -g/2 t^2+ v_0 sin(/theta)t ?
 
I suppose it's mostly the /2t^2 I'm failing to see...
 
What are the forces along the plane... what are the forces perpendicular to the plane...
 
i see that gravity is affecting v(y), but why /2t^2?
 
Oblio said:
i see that gravity is affecting v(y), but why /2t^2?

that's just for the (1/2)gt^2.

Hint: Rotate your sketch... so that the incline is horizontal... now you can treat it like a regular projectile problem... the only differences are the vertical and horizontal accelerations...
 
  • #10
learningphysics said:
that's just for the (1/2)gt^2.

Where does that come from?
 
  • #11
Oblio said:
Where does that come from?

From the kinematics equation d = v0*t + (1/2)at^2

a = -g.
 
  • #12
When the projectile is in the air... the only force acting on it is gravity... what is the component of gravity along the plane... what is the component of gravity perpendicular to the plane...

so what's the component of acceleration along the plane... what's the component of acceleration perpendicular to the plane...
 
  • #13
HallsofIvy said:
You can solve the first equation for t: t= x/(v_0 cos(\theta) and put that into the equation for y: y= -g/(2v_0 cos(\theta)x^2+ (sin(\theta)/cos(\theta))x to get the equation of the parabola. Now it is a question of determing where that parabola intersects the line giving the inclined plane. That is, of course, y= tan(\phi)x= (sin(\phi)/cos(\phi))x.

Sorry, duh. 'Brainfart' there Learningphysics.
I more or less understand the steps, but could you explain the logic behind why you put the equation for x into y?

And why is 'y= tan(\phi)x= (sin(\phi)/cos(\phi))x ' where the parabola interesets?
 
  • #14
Oblio said:
Sorry, duh. 'Brainfart' there Learningphysics.
I more or less understand the steps, but could you explain the logic behind why you put the equation for x into y?

And why is 'y= tan(\phi)x= (sin(\phi)/cos(\phi))x ' where the parabola interesets?

That's HallofIvy's post, not mine. The thing is Halls is considering x along the horizontal... y along the vertical... I think the question wants you to approach the problem a little differently...

Work the problem taking x to be along the plane... y to be perpendicular to the plane... I think it makes the problem a lot easier...
 
  • #15
That it what i was trying to do...

so don't do it the way halls of ivy said? I am not sure where to go with this now...
 
  • #16
Oblio said:
That it what i was trying to do...

so don't do it the way halls of ivy said? I am not sure where to go with this now...

See my previous post about rotating the axes, so that the incline is horizontal...
 
  • #17
yeah that is what i had done originally.
 
  • #18
Oblio said:
yeah that is what i had done originally.

cool. can you show how far you got?
 
  • #19
not very. Not using HallsofIvy's method I'm not sure how to figure the distance and parabolic motion of the ball.
 
  • #20
Oblio said:
not very. Not using HallsofIvy's method I'm not sure how to figure the distance and parabolic motion of the ball.

Given v0 and that the angle at which the ball is launched above the incline is theta... what is the initial velocity along the incline... what is the initial velocity perpendicular to the incline ?
 
  • #21
ok, so i still separate THAT into its x and y coordinates.

vx = vocos(theta) and vy=vosin(theta)
 
  • #22
Oblio said:
ok, so i still separate THAT into its x and y coordinates.

vx = vocos(theta) and vy=vosin(theta)

exactly... now what is the acceleration in the x direction (along the plane)... what is the acceleration in the y direction (perpendicular to the plane) ?
 
  • #23
ay = -g and ax = 0 i believe.
 
  • #24
Oblio said:
ay = -g and ax = 0 i believe.

remember... we're taking x to be along the plane... y to be perpendicular to the plane...
 
  • #25
Oblio said:
ok, so i still separate THAT into its x and y coordinates.

vx = vocos(theta) and vy=vosin(theta)

Yes. I want you to remember this is vx along the plane... vy perpendicular to the plane...

The angle at which which the ball is launched relative to the horizontal is (theta + fi)... so if we were working with the x as the horizontal and y as the vertical it would be:

vx = vocos(theta + fi) and vy=vosin(theta + fi)

But since we're working with x along the plane... y perpendicular to the plane... it is:

vx = vocos(theta) and vy=vosin(theta)
 
  • #26
ay=-gcos(phi) ?
 
  • #27
Oblio said:
ay=-gcos(phi) ?

exactly. And what is ax ?
 
  • #28
ahhh ax=-gsin(phi)
 
  • #29
will i be setting vy to zero and then solving a free fall?
 
  • #30
Oblio said:
ahhh ax=-gsin(phi)

yes, exactly...

So what is the displacement in the y direction... what is the displacement in the x direction?
 

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