A ball thrown at an angle above an inclined plane

  • #1
Homework Statement:
Suppose a ball is thrown straight up with an initial speed of ##v_i## on an inclined plane, where the plane is inclined at an angle of ##\phi## above the horizontal and the ball's initial velocity is at an angle ##\theta## above the plane.
Relevant Equations:
##F=ma##
I understand how the diagram below determined the ##x## and ##y## axis for the velocity vectors but I don't understand the gravity vectors. What I don't understand about the gravity vectors is why is ##-mg## in the ##y-##axis equal to ##-mg\cos\theta## and the ##x-##axis is equal to ##-mg\cos\theta## even though the diagram for the velocity vector has the ##y-##axis as ##v_0\sin\theta## which is exactly on the same axis for the gravity vector so shouldn't the gravity vectors ##\cos## and ##\sin## be flipped? Also, I don't understand how the ##\phi## angle below the horizontal was determined in the diagram. Shouldn't that ##\phi## be instead ##\theta##?

IMG_1758.jpg
 

Answers and Replies

  • #2
kuruman
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Homework Statement:: Suppose a ball is thrown straight up with an initial speed of ##v_i## on an inclined plane, where the plane is inclined at an angle of ##\phi## above the horizontal and the ball's initial velocity is at an angle ##\theta## above the plane.
Relevant Equations:: ##F=ma##

I understand how the diagram below determined the ##x## and ##y## axis for the velocity vectors but I don't understand the gravity vectors. What I don't understand about the gravity vectors is why is ##-mg## in the ##y-##axis equal to ##-mg\cos\theta## and the ##x-##axis is equal to ##-mg\cos\theta## even though the diagram for the velocity vector has the ##y-##axis as ##v_0\sin\theta## which is exactly on the same axis for the gravity vector so shouldn't the gravity vectors ##\cos## and ##\sin## be flipped? Also, I don't understand how the ##\phi## angle below the horizontal was determined in the diagram. Shouldn't that ##\phi## be instead ##\theta##?

View attachment 269459
There is no ##mg\cos\theta## in the diagram. The weight ##mg## down is correctly resolved into components where the ##x##-axis is up the incline and the ##y##-axis perpendicular and away from the incline and ##\phi## is the angle of the incline. Note that ##\theta## is complementary to ##\phi## so that ##\cos\phi= \sin\theta## and ##\sin\phi=\cos\theta.##

Angle ##\phi## below the horizontal is equal to angle ##phi## above the horizontal because they have their sides mutually perpendicular. That's a theorem from geometry. You can see why that is if you imagine the angle ##phi## above the horizontal being decreased to zero in which case angle ##\phi## below the horizontal will correspondingly go to zero.
 
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  • #3
There is no ##mg\cos\theta## in the diagram. The weight ##mg## down is correctly resolved into components where the ##x##-axis is up the incline and the ##y##-axis perpendicular and away from the incline and ##\phi## is the angle of the incline. Note that ##\theta## is complimentary to ##\phi## so that ##\cos\phi= \sin\theta## and ##\sin\phi=\cos\theta.##

Angle ##\phi## below the horizontal is equal to angle ##phi## above the horizontal because they have their sides mutually perpendicular. That's a theorem from geometry. You can see why that is if you imagine the angle ##phi## above the horizontal being decreased to zero in which case angle ##\phi## below the horizontal will correspondingly go to zero.

Ops, I meant to say ##mg\cos\phi## and ##mg\sin\phi##, not ##mg\cos\theta## and ##mg\sin\theta## . Why are they complimentary even though that in the diagram is clear that ##v_0\sin\theta## and ##-mg\cos\phi## are on the same axis so shouldn't they both have the same trig function of ##\sin##?

Also, do you know of the name of the geometrical theorem?
 
  • #4
kuruman
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Ops, I meant to say ##mg\cos\phi## and ##mg\sin\phi##, not ##mg\cos\theta## and ##mg\sin\theta## . Why are they complimentary even though that in the diagram is clear that ##v_0\sin\theta## and ##-mg\cos\phi## are on the same axis so shouldn't they both have the same trig function of ##\sin##?

Also, do you know of the name of the geometrical theorem?
Complementary means ##\theta +\phi=90^o.## It sure looks like it in your diagram.

Screen Shot 2020-09-15 at 4.27.32 PM.png

As I said before, ##v_0\sin\theta=v_0\cos\phi## because the angles are complimentary, so they do have the same trig function.

I do not think that the theorem has a name. It was proven by Euclid who lived 23 centuries ago. It suffices to say that "two acute angle that have their sides mutually perpendicular are equal." It's a good one to remember.
 
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  • #5
Lnewqban
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What is the reason for the inclined plane to be?
The trajectory of the ball is going to be perfectly vertical, regardless the angles shown.
Is this part of a problem that has more questions or conditions?
 
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  • #6
kuruman
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What is the reason for the inclined plane to be?
The trajectory of the bal is going to be perfectly vertical, regardles the angles shown.
Is this part of a problem that has more questions or conditions?
Only part of the problem was provided, not the whole thing or the question to be addressed, therefore I cannot speculate about the inclined plane's raison d'être.
 
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  • #7
Ty @kuruman .

@Lnewqban it is just part of the problem and not the whole question. I just needed clarificaiton of the diagram which is why I didn't write the whole question.
 
  • #8
Lnewqban
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Ty @kuruman .

@Lnewqban it is just part of the problem and not the whole question. I just needed clarificaiton of the diagram which is why I didn't write the whole question.
Thank you. :smile:
 
  • #9
haruspex
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Homework Statement:: Suppose a ball is thrown straight up with an initial speed of ##v_i## on an inclined plane, where the plane is inclined at an angle of ##\phi## above the horizontal and the ball's initial velocity is at an angle ##\theta## above the plane.
Is this a translation? I suspect it is thrown 'straight up' in the sense that its trajectory is in a verical plane orthogonal to the plane of the slope. It is thrown at an angle θ+φ to the horizontal, not vertically.
 
  • #10
Is this a translation? I suspect it is thrown 'straight up' in the sense that its trajectory is in a verical plane orthogonal to the plane of the slope. It is thrown at an angle θ+φ to the horizontal, not vertically.
Yes, correct. I am unable to edit the question so I am unable to add to it but I should add that ##\theta > \phi##.
 
  • #11
haruspex
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Yes, correct. I am unable to edit the question so I am unable to add to it but I should add that ##\theta > \phi##.
No, it's not that ##\theta > \phi##. That may or may not be true. It's that ##\theta + \phi\leq \frac{\pi}2##.
 
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