A ball thrown at an angle above an inclined plane

In summary, the conversation discusses the diagram of an inclined plane and a ball being thrown straight up with initial velocity. The gravity vectors are resolved into components and the angles of the incline and initial velocity are complementary. The reason for the inclined plane is not specified.
  • #1
tryingtolearn1
58
5
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
 
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  • #2
tryingtolearn1 said:
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
kuruman said:
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
tryingtolearn1 said:
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
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
Lnewqban said:
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
tryingtolearn1 said:
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
tryingtolearn1 said:
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.
 
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  • #10
haruspex said:
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
tryingtolearn1 said:
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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Related to A ball thrown at an angle above an inclined plane

1. How does the angle at which the ball is thrown affect its trajectory on an inclined plane?

The angle at which the ball is thrown will determine the initial velocity and direction of the ball. This will affect the trajectory of the ball on the inclined plane, as a higher angle will result in a steeper initial path and a lower angle will result in a more gradual path.

2. What factors besides angle can affect the motion of a ball thrown at an inclined plane?

The mass and initial velocity of the ball, as well as the angle of the inclined plane, can also affect the motion of the ball. Friction and air resistance may also play a role in the ball's trajectory.

3. How does the height of the inclined plane impact the motion of the ball?

The height of the inclined plane will affect the distance the ball travels before hitting the ground. A higher inclined plane will result in a longer distance traveled, while a lower inclined plane will result in a shorter distance.

4. Can the ball ever reach the top of the inclined plane?

It is possible for the ball to reach the top of the inclined plane if it has enough initial velocity and the angle of the inclined plane is steep enough. However, factors such as friction and air resistance may prevent the ball from reaching the top.

5. How does the mass of the ball affect its motion on an inclined plane?

The mass of the ball will affect the force of gravity acting on it, which will in turn affect its acceleration and velocity. A heavier ball will require more force to move up the inclined plane, and may also experience more friction and air resistance.

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