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.

  • #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...
 
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  • #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.
 
  • #51
oops! little pen and ink mistake over here..

t = vosin(theta) / (1/2)gcos(phi)
yep!
 
  • #52
Oblio said:
oops! little pen and ink mistake over here..

t = vosin(theta) / (1/2)gcos(phi)
yep!

cool. plugging that into dx should work.
 
  • #53
dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - (1/2)gsin(phi)vosin(theta)/(1/2)gcos(phi)
 
  • #54
Oblio said:
dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - (1/2)gsin(phi)vosin(theta)/(1/2)gcos(phi)

you missed t^2... didn't square t.
 
  • #55
oops.
man I am bad..
 
  • #56
dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - (1/2)gsin(phi)(vosin(theta)/(1/2)gcos(phi))^2

on the right side i can cancel out the (1/2) on the bottom and top, as well as the g, giving,

dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - sin(phi)(vosin(theta)/cos(phi))^2
 
  • #57
Oblio said:
dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - (1/2)gsin(phi)(vosin(theta)/(1/2)gcos(phi))^2

on the right side i can cancel out the (1/2) on the bottom and top, as well as the g, giving,

dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - sin(phi)(vosin(theta)/cos(phi))^2

careful the 1/2 and g are squared...
 
  • #58
oops again.

so I am left with
dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - sin(phi)(vo^2 sin(theta) ^2 / (1/2) g (cos(phi)^2)
 
  • #59
Oblio said:
oops again.

so I am left with
dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - sin(phi)(vo^2 sin(theta) ^2 / (1/2) g (cos(phi)^2)

Looks good... try to simplify and use a trig identity to get it to look like the formula they gave for the range...
 
  • #60
will trig identities apply since theta and phi are present?
 

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