Titled reference frame, N2L with position and velocity

AI Thread Summary
In this discussion, participants analyze the motion of a ball thrown up an inclined plane, focusing on applying Newton's second law and deriving the ball's trajectory. They clarify that the ball's initial velocity components can be determined using trigonometric functions based on the launch angle relative to the incline. The conversation emphasizes the need to express the ball's position over time and find the intersection with the inclined plane to calculate the range. Key formulas for the range and maximum range are derived, with participants discussing the correct application of kinematic equations and the effects of gravity on the motion. Ultimately, the discussion aims to clarify the problem-solving approach for projectile motion on an incline.
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?
 
  • #31
Oblio said:
will i be setting vy to zero and then solving a free fall?

I don't understand... the question asks you to find the range in the x direction I believe...
 
  • #32
using kinematics?
 
  • #33
Oblio said:
using kinematics?

yes... what is the displacement along the x direction and y direction... in terms of time...
 
  • #34
d = v0*t + (1/2)at^2
 
  • #35
Oblio said:
d = v0*t + (1/2)at^2

Yes. Now use the values we've calculated for v0 and a for this problem... for the x- direction... and the y-direction...
 
  • #36
dx = vocos\varthetat + (1/2)(-g)t^2sin\phi

dy = vosin\varthetat + (1/2)(-g)t^2cos\phi
 
Last edited:
  • #37
i have a feeling this is wrong for dy...
 
Last edited:
  • #38
Oblio said:
i have a feeling this is wrong for dy...

Looks right to me. So now you want to find the range...
 
  • #39
range?
 
  • #40
Oblio said:
range?

the question asks for the range...
 
  • #41
which is really just the displacement up the plane?
 
  • #42
Oblio said:
which is really just the displacement up the plane?

yes, maximum displacement... the object is thrown at an angle, then hits the incline eventually... the displacement up the plane when it hits is the range...
 
  • #43
is this the same kinematics equation?
 
  • #44
Oblio said:
dx = vocos\varthetat + (1/2)(-g)t^2sin\phi

dy = vosin\varthetat + (1/2)(-g)t^2cos\phi

use these equations to find dx when dy = 0... that gives the range.
 
  • #45
solve for... vo? and insert into dx?
 
  • #46
Oblio said:
solve for... vo? and insert into dx?

no, you want to eliminate t... so solve for t and substitute into dx.
 
  • #47
without simplifying yet..

dx = vocos (theta) (vosin (theta) / (-1/2)gsin(phi)) + (1/2)(-g)cos(phi)((vosin(theta)/(-1/2)gsin)^2)

ya?
 
  • #48
Oblio said:
without simplifying yet..

dx = vocos (theta) (vosin (theta) / (-1/2)gsin(phi)) + (1/2)(-g)cos(phi)((vosin(theta)/(-1/2)gsin)^2)

ya?

hmm... close but some mistakes... can you post what you got for t?
 
  • #49
t = vosin(theta) / (-1/2)gsin(phi)
 
  • #50
Oblio said:
t = vosin(theta) / (-1/2)gsin(phi)

setting dy = 0, and solving for t will give you something different... it should be cos(phi) and the - shouldn't be there.
 

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