How Does Projectile Angle Affect Asymptotic Behavior in Physics Graphs?

In summary, the projectile equation states that the horizontal velocity must be 0 for it to hit the ground perpendicularly. After solving for t, it can be concluded that t is equal to the distance from the launch point to the impact point.
  • #1
SayedD
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Homework Statement
A particle is shot perpendicular to respect to the inclined plane with inclination α, with initial velocity v. Determine the value of v so that the following particle hit the incline plane with inclination β perpendiculrly. The distance between the lowest point to the launching point is a.

I have been struggling all day trying to solve this by using a method which involves tilting the ground so it makes an inclined plane with inclination α+β, here are my steps
Relevant Equations
Given in my attempt
So using the projectile equation I got ##a\sin (\alpha + \beta) = v\sin (\alpha + \beta)t + \frac{g\sin \beta t^2}{2}##

To find t we use the other projectile equation:
Since the horizontal velocity must be 0 for it to hit the ground perpendicularly we set the equation to be:
##0 = v\sin (\alpha + \beta) - g\sin \beta t## \\
##t = \frac{v\sin (\alpha + \beta)}{g\sin \beta}##

please help solve using this method
 

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  • #2
What other condition must be satisfied when the projectile hits?
 
  • #3
The acceleration along your new y direction is not g, is it?
 
  • #4
Is this a multiple choice question with the correct answer being one of A - E in the third photo? If so, then I see two courses of action
Plan A: Solve the problem correctly and compare your answer to what is given.
Plan B: Solve the problem for a special choice of ##\alpha## and ##\beta## and see which of the given five answers reduces to what you know to be the case.

I went with plan B and I chose ##\alpha = \beta = 45^o.## Then,
(a) the projection and landing angles relative the horizontal are also 45°;
(b) the trajectory is symmetric about the point where the inclines meet;
(c) the projectile returns to the same height from which it is launched;
(d) the horizontal range of the projectile is ##R=\dfrac{v_0^2}{g}##;
(e) the horizontal range is also ##R=2\times a \cos(45^o)=\sqrt{2}a##.

It follows, then, that when ##\alpha = \beta = 45^o##, ##v_0^2=\sqrt{2}ag.##

Which of the five choices gives this result? I leave that up to you to figure out.
 
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  • #5
@kuruman that is a clever approach. I have started a stab at doing Plan A and I'm not sure if I am on the right track yet or not. It is very easy to make an algebra mistake along the way. And looking at choices A - E, quite a bit of algebra will go into this.

@SayedD what language is the third image in? I have a translator app but it isn't making any headway on that problem statement.
 
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  • #6
gmax137 said:
@kuruman that is a clever approach. I have started a stab at doing Plan A and I'm not sure if I am on the right track yet or not. It is very easy to make an algebra mistake along the way. And looking at choices A - E, quite a bit of algebra will go into this.

@SayedD what language is the third image in? I have a translator app but it isn't making any headway on that problem statement.
I doubt that any of the answers will match the special case I chose. Look at what happens to the denominator when both tangents are equal to 1. I will try plan A at some point because this is one projectile motion problem I have not seen before. The algebra and the strategy can be simplified with the shortcut equations (4) and (5) in this insight that I put together a few months back.

Google translate tells me that "Sebuah benda bermassa m dilempar tegak lurus terhadap ##\dots~##" is Indonesian for "An object of mass ##m## is thrown perpendicular to ##\dots~##"
 
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  • #7
The denominator under the square root sign looks a bit suspect. It's going to be negative presuming ##\alpha## and ##\beta## are positive. So then g would have to be taken as negative otherwise the root is non-real ?

Per Kuruman's suggestion , one solution should be correct when ##\alpha=\beta=45^{\circ}##.
 
  • #8
SayedD said:
Homework Statement: A particle is shot perpendicular to respect to the inclined plane with inclination α, with initial velocity v. Determine the value of v so that the following particle hit the incline plane with inclination β perpendicularly. The distance between the lowest point to the launching point is a.

I have been struggling all day trying to solve this by using a method which involves tilting the ground, so it makes an inclined plane with inclination α+β, ...
Welcome, SayedD!
Could you verify that the "lowest point" is the correct translation?
If so, what would the meaning of that point would be, other than the intersection of both surfaces, each of those being perpendicular to the trajectory?
I believe that tilting the planes and the direction of gravity can't simplify things in this case.
 
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  • #9
Lnewqban said:
Welcome, SayedD!
Could you verify that the "lowest point" is the correct translation?
I wonder if a is supposed to be the direct distance between launch point and impact point ?
 
  • #10
the variables are shown clearly in the second (middle) image; and it appears to be typeset rather than hand-drawn (so I take it to be correct and not a mis-interpretation). That said, I have still not solved this one. I hope to get some time later today.

I use these homework exercises as little mind puzzles, after being out of school for decades.
 
  • #11
Gravitational torque = rate of change of angular momentum

##\vec{a}\times \vec{s}=\vec{v}\times \vec{u}##

Here is the calculation on WA.

It does not look at all like one of the given selection but it does pass the "Kuruman test" with ##\alpha=\beta=45^{\circ}##. For a projectile launched with velocity v at 45 degrees, maximum horizontal range ##v^2/g## is achieved and that works out to ##\sqrt{2} a## as expected. Note I have used p and q for the angles on WA

Here is the "Kuruman test" calculation in which I have replaced p and q with ##45^{\circ}##.
 
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  • #12
I picked the angles to be equal to 45° because that makes the evaluation of each choice trivial since the tangents are equal to 1. One can easily see that the identical denominators turn negative when the angles are equal.

Here is a more general test for equal angles ##\alpha=\beta##. The projection angle ##\theta## relative to the horizontal is ##\theta=\frac{\pi}{2}-\alpha##. The projectile returns to the same height from which it was launched in which case the horizontal range is given by $$R=\frac{2v_0^2\sin\!\theta\cos\!\theta}{g}=\frac{2v_0^2\cos\!\alpha\sin\!\alpha}{g}.$$ The horizontal range is also twice the projection of length ##a## on the horizontal axis, ##R=2\times a\cos\!\alpha##. Thus,$$\frac{2v_0^2\cos\!\alpha\sin\!\alpha}{g}= 2a\cos\!\alpha\implies v_0^2=\frac{ag}{\sin\!\alpha}.$$It follows that any expression for the initial speed must be such that ##v_0^2(\alpha, \alpha,a)=\dfrac{ag}{\sin\!\alpha}.##
It also follows that ##v_0^2(\alpha, \frac{\pi}{2},a)=\dfrac{ag}{\sin\!\alpha}.## Because of the symmetry of the trajectory, the projectile reaches maximum height when it is above the point of intersection of the two planes and is moving horizontally. This means that at that point it will hit a vertical wall (##\beta =\frac{\pi}{2}##) perpendicularly.

Thus, any expression ##v_0^2(\alpha, \beta,a)## must reduce to ##\frac{ag}{\sin\!\alpha}## when ##\alpha =\beta## and when ##\beta=\frac{\pi}{2}##. I have obtained such an expression, but it looks ghastly so I am trying to make it look prettier before I share it. Stay tuned.
 
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  • #13
kuruman said:
Thus, any expression ##v_0^2(\alpha, \beta,a)## must reduce to ##\frac{ag}{\sin\!\alpha}## when ##\alpha =\beta## and when ##\beta=\frac{\pi}{2}##. I have obtained such an expression, but it looks ghastly so I am trying to make it look prettier before I share it. Stay tuned.
The expression I obtained above from WA does exactly that.

1647614242298.png
 
  • #14
I should add thanks for providing 'test bed' conditions for the equation!
 
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  • #15
neilparker62 said:
The expression I obtained above from WA does exactly that.

View attachment 298575
And agrees with mine with ##p=\alpha## and ##q=\beta.##
 
  • #16
Now we just need to figure if this expression is equivalent to any of A to E in the OP's question.
 
  • #17
Option A passes the "improved Kuruman test": ##{v_0}^2=\frac{ag}{\sin{\alpha}}## when ##\alpha=\beta##. Except that the sign is wrong. It fails the test with ##\beta=90^{\circ}## claiming 'undefined' presumably because it attempts to evaluate ##\tan{90^{\circ}}##
 
  • #18
Here is what I found after some investigation.

My ghastly expression is $$\begin{align}v_{\text{0A}}^2=\frac{2 \sin (\alpha +\beta )}{2 \sin (\alpha ) \sin (\alpha +\beta )-\frac{\cos (\beta ) \left(\sin ^2(\beta )-\sin ^2(\alpha )\right)}{\sin ^2(\beta )}}.\end{align}$$ Your (neater) expression is $$\begin{align} v_{\text{0B}}^2=\frac{4 \sin ^2(\beta )}{3 \sin (\alpha )-\sin (\alpha +2 \beta )}.\end{align}$$ The expression in choice A with the denominator multiplied by -1 to make it positive when ##\alpha=\beta~## is $$\begin{align}v_{\text{0C}}^2=\frac{\cos (\alpha ) \left(2 \tan ^2(\alpha ) \tan ^2(\beta ) (\tan (\alpha )+\tan (\beta ))\right)}{\sin ^2(\alpha ) \left(2 \tan (\alpha ) \tan ^3(\beta )+2 \tan ^2(\alpha ) \tan ^2(\beta )-\tan ^2(\alpha )+\tan ^2(\beta )\right)}.\end{align}$$
I. Equations (1) and (2) give identical results but equation (3) does not. I verified this by fixing ##\alpha## to some value and plotting ##v0^2## as a function of ##\beta## from ##0## to ##\pi/2##. The graphs for (1) and (2) were on top of each other whilst the graph for (3) was not. From now on I will refer to equations (1) and (2) collectively as "our equation" and equation (3) as "their equation" by replacing sines and cosines with tangents.

II. The condition for our equation to be positive is $$\tan(\alpha)>\frac{\sin(\beta)\cos(\beta)}{1+\sin^2(\beta)}.$$The ratio on the right has a maximum value of 0.3535 which translates to ##\alpha > 19.5^o.## In other words, as long as the above condition is met, our equation will always have a solution. This is not true for their equation which changes sign in the region ##0<\beta<\pi/2## no matter what one chooses for ##\alpha##. I verified this last result by doing several plots. The one below is for ##\alpha = 21.6^o##. The blue line is our equation and the orange line is their equation. The vertical line marks the value of ##\alpha##. It is ironic and instructive to see that our equation matches theirs only at ##\alpha=\beta## and ##\beta=\frac{\pi}{2}##, i.e. both equations pass the tests that I proposed in post #12.

AlphaBeta.png

III. Both our and their equations agree when ##\alpha = \beta## and when ##\beta = \frac{\pi}{2}.## One can get around the "undefined" problem with their equation (3) if one divides both numerator and denominator by ##\tan^3(\beta)## and then take the limit as ##\tan(\beta)\rightarrow \infty.##

Edits: After some algebra, I was able to transform equation (1) into equation (2) so they are identical. I wonder if it is worth anyone's while to see how close one can get to their equation.

Also edited to add the vertical line at ##\beta =\alpha## in the plot.

Final edit: Equations (1) and (2) can be algebraically transformed to one of the options with the negative signs in front of the factors of ##2## in the denominator flipped to positive signs.
 
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  • #20
Nice. Can you expand this to two sliders that will vary the slopes independently? Sorry, for being greedy.
 
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  • #22
This is lovely! Thank you for doing this.
 
  • #23
Pleasure - thanks also to @robphy who introduced me to sliders in Demos graphs. Always good to have a 'real' problem to work on when you are learning something new!
 
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  • #24
Given: $$v_{\text{0B}}^2=\frac{4ag \sin ^2(\beta )}{3 \sin (\alpha )-\sin (\alpha +2 \beta )},$$ the graph of ##v_{\text{0B}}## vs ##\beta## will have an asymptote if ##3\sin\alpha \leq 1 \implies \alpha \leq 19.47^{\circ}##.

In the following desmos graph, ##\alpha## is slider controlled and ##v_{0}(\alpha,\beta)## vs ##\beta## is plotted for ##0^{\circ} \leq \beta \leq 90^{\circ}##. Asymptotic behaviour is clear when ##\alpha## drops below the above limit - otherwise the graph obtained is a continuous function over the given domain.

https://www.desmos.com/calculator/s9b8pwir3u
 
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Related to How Does Projectile Angle Affect Asymptotic Behavior in Physics Graphs?

1. What is an Olympiad Projectile Problem?

An Olympiad Projectile Problem is a type of physics problem that involves the motion of a projectile, such as a ball or a bullet, in a two-dimensional space. These problems often require the use of kinematic equations and trigonometry to solve.

2. How do I approach solving an Olympiad Projectile Problem?

The first step in solving an Olympiad Projectile Problem is to draw a diagram of the problem and label all known and unknown quantities. Then, use the given information to set up equations using the kinematic equations and solve for the unknown variable.

3. What are some common strategies for solving Olympiad Projectile Problems?

Some common strategies for solving Olympiad Projectile Problems include breaking the problem down into smaller parts, using symmetry to simplify the problem, and using trigonometric identities to eliminate variables.

4. What are some common mistakes to avoid when solving Olympiad Projectile Problems?

Some common mistakes to avoid when solving Olympiad Projectile Problems include not labeling the diagram correctly, using incorrect units, and not considering air resistance or other external forces.

5. How can I improve my skills in solving Olympiad Projectile Problems?

To improve your skills in solving Olympiad Projectile Problems, practice solving a variety of problems and familiarize yourself with different strategies and techniques. It is also helpful to review the basic principles of projectile motion and kinematic equations.

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