A ball thrown at an angle above an inclined plane

[tryingtolearn1]

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 $\theta$?

 

[kuruman]

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.

 

[tryingtolearn1]

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?

 

[kuruman]

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.

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.

 

[Lnewqban]

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?

 

[kuruman]

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.

 

[tryingtolearn1]

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.

 

[Lnewqban]

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.

 

[haruspex]

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.

 

[tryingtolearn1]

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 $\theta > \phi$.

 

[haruspex]

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 $\theta + \phi\leq \frac{\pi}2$.