Projectile motion with only distance, an angle and a ratio

In summary, A projectile with an initial launch angle of θ=41.86° and a distance to the wall of d=11.06 m strikes the wall with a speed that is a fraction f=0.925 of the initial speed. To find the initial velocity, the equation v=fu can be used, along with the equations v_x=u_x=u*cos(θ) and v_y=u_y-gt=u*sin(θ)-gt. The value of d does not directly factor into these equations, so further information or equations may be needed to solve for the initial velocity.
  • #1
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Homework Statement


launch angle of θ=41.86°. The distance to the wall is d=11.06 m, and it strikes the wall with speed that is a fraction f=0.925 of the initial speed, while it is still gaining height (i.e., it has not the highest point of the trajectory). What is the initial velocity? Express your answer in m/s.

Homework Equations

The Attempt at a Solution


V(initial)=v(final)/0.925
 
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  • #2
Julie Yum said:

Homework Statement


launch angle of θ=41.86°. The distance to the wall is d=11.06 m, and it strikes the wall with speed that is a fraction f=0.925 of the initial speed, while it is still gaining height (i.e., it has not the highest point of the trajectory). What is the initial velocity? Express your answer in m/s.

Homework Equations

The Attempt at a Solution


V(initial)=v(final)/0.925

Good morning from CET (+DST). According to this forum's philosophy, your goal should be to learn to solve this kind of problems and to find the answer yourself (with assistance of a more experienced member). Therefore you are expected to show effort to solve the problem, so please let us know what you've achieved so far, what formulas/equations/principles you think you can use and describe with which of the steps of the solution you have problems with. So what are your ideas?
 
  • #3
@Julie Yum this looks like quite a difficult problem. It will test your algebra and your mathematical understanding of the equations of projectile motion.
 
  • #4
PeroK said:
@Julie Yum this looks like quite a difficult problem. It will test your algebra and your mathematical understanding of the equations of projectile motion.
stockzahn said:
Good morning from CET (+DST). According to this forum's philosophy, your goal should be to learn to solve this kind of problems and to find the answer yourself (with assistance of a more experienced member). Therefore you are expected to show effort to solve the problem, so please let us know what you've achieved so far, what formulas/equations/principles you think you can use and describe with which of the steps of the solution you have problems with. So what are your ideas?

So far I have created a multiple formulas to relate the problem to one variable, but I am not sure if I am doing it correctly. I have included a link to the image with my work so far. Please let me know if I am on the right track, or if I should start totally over.
https://drive.google.com/open?id=1IOu8GFxsn8uUyy96XoYpBVFcTIGmX6Op
 
  • #5
stockzahn said:
Good morning from CET (+DST). According to this forum's philosophy, your goal should be to learn to solve this kind of problems and to find the answer yourself (with assistance of a more experienced member). Therefore you are expected to show effort to solve the problem, so please let us know what you've achieved so far, what formulas/equations/principles you think you can use and describe with which of the steps of the solution you have problems with. So what are your ideas?
Here is a link to what I have so far: https://drive.google.com/open?id=1IOu8GFxsn8uUyy96XoYpBVFcTIGmX6Op
What I have seems very complex, long and hard to describe using a keyboard type format, so I excluded it from the forum. However, I have found a way to show what I have as a link. I would appreciate any feedback if possible.
 
  • #6
Julie Yum said:
So far I have created a multiple formulas to relate the problem to one variable, but I am not sure if I am doing it correctly. I have included a link to the image with my work so far. Please let me know if I am on the right track, or if I should start totally over.
https://drive.google.com/open?id=1IOu8GFxsn8uUyy96XoYpBVFcTIGmX6Op

This problem is not really suited to plugging in all those numbers. For one thing, it's almost impossible to tell whether you are on the right track. It's just a mass of numbers in which the actual physics is buried.

I find ##u, v## easier than ##v_0, v_1##. In any case, you should be looking at an algebraic approach, with ##d, f, \theta, t## as the variables.

The first step is to write down what you are given. I would start with:

##v = fu##, hence ##v^2 = f^2u^2##

##v_x = u_x = u\cos\theta##,
##v_y = u_y - gt = u\sin\theta - gt##

Now, can you continue from there? Where does ##d## come in?
 
  • #7
PeroK said:
This problem is not really suited to plugging in all those numbers. For one thing, it's almost impossible to tell whether you are on the right track. It's just a mass of numbers in which the actual physics is buried.

I find ##u, v## easier than ##v_0, v_1##. In any case, you should be looking at an algebraic approach, with ##d, f, \theta, t## as the variables.

The first step is to write down what you are given. I would start with:

##v = fu##, hence ##v^2 = f^2u^2##

##v_x = u_x = u\cos\theta##,
##v_y = u_y - gt = u\sin\theta - gt##

Now, can you continue from there? Where does ##d## come in?

What does u and f represent in the formula?
 
  • #8
Julie Yum said:
What does u and f represent in the formula?

##u## is the initial speed. ##f## is given in the question you posted.

You can use ##v_0## instead of ##u##, but I find ##u## easier.
 
  • #9
PeroK said:
##u## is the initial speed. ##f## is given in the question you posted.

You can use ##v_0## instead of ##u##, but I find ##u## easier.
Here is a link to my work, as you have advised me, please let me know if this seems like the direction i should head.
https://drive.google.com/open?id=1jzoSRc4l44SapTDsZS1_4LWnbba1O50M
 
  • #10
Julie Yum said:
Here is a link to my work, as you have advised me, please let me know if this seems like the direction i should head.
https://drive.google.com/open?id=1jzoSRc4l44SapTDsZS1_4LWnbba1O50M

Your calculation for ##t## is not correct. ##d## is a horizontal distance.
 
  • #11
PeroK said:
Your calculation for ##t## is not correct. ##d## is a horizontal distance.

Would this seem more like it?
https://drive.google.com/open?id=1TEuQbkn5YkszZbLTtcFelfTiPgDzVWFG
 
  • #12
Julie Yum said:
Would this seem more like it?
https://drive.google.com/open?id=1TEuQbkn5YkszZbLTtcFelfTiPgDzVWFG

The first equation looks better now:

##v_y = u\sin\theta - \frac{gd}{u\cos\theta}##

Although, as you can see, I would prefer ##gd## to ##108.4986##!

Now, you need another equation in ##v_y##. Remember that ##v^2 = v_x^2 + v_y^2##.

... actually, you are nearly there with that second equation.
 
Last edited:
  • #13
PeroK said:
The first equation looks better now:

##v_y = u\sin\theta - \frac{gd}{u\cos\theta}##

Although, as you can see, I would prefer ##gd## to ##108.4986##!

Now, you need another equation in ##v_y##. Remember that ##v^2 = v_x^2 + v_y^2##.

... actually, you are nearly there with that second equation.
Any tips on how to isolate for u?
https://drive.google.com/open?id=18knkzNBXae8VPcX9df6Q52TLjFJ105pB
 
  • #14
Julie Yum said:
Any tips on how to isolate for u?
https://drive.google.com/open?id=18knkzNBXae8VPcX9df6Q52TLjFJ105pB

Yes, a better approach is:

##v_x^2 + v_y^2 = f^2u^2##
##v_y^2 = f^2u^2 - u^2\cos^2\theta##

Which leads to another equation for ##v_y##. You can then equate this with the equation for ##v_y## you already have.
 
  • #15
Julie Yum said:
Here is a link to what I have so far: https://drive.google.com/open?id=1IOu8GFxsn8uUyy96XoYpBVFcTIGmX6Op
What I have seems very complex, long and hard to describe using a keyboard type format, so I excluded it from the forum. However, I have found a way to show what I have as a link. I would appreciate any feedback if possible.

You really should type it out here; most helpers (me included) will not open such files and so will not look at your work. Read the post "Guidelines for students and helpers" for more on this issue.

You do not need to type every single step; you can just type summaries along the way, so you can say things like "using the first two equations we get..." and "solving the above equation we find that ... ", inserting the solution formulas, but. If you use LaTeX, that is all quite easy; see, https://www.physicsforums.com/help/latexhelp/
 
  • #16
PeroK said:
Yes, a better approach is:

##v_x^2 + v_y^2 = f^2u^2##
##v_y^2 = f^2u^2 - u^2\cos^2\theta##

Which leads to another equation for ##v_y##. You can then equate this with the equation for ##v_y## you already have.

How would I continue to isolate for u?
https://drive.google.com/open?id=1IOu8GFxsn8uUyy96XoYpBVFcTIGmX6Op
 
  • #17
Julie Yum said:
How would I continue to isolate for u?
https://drive.google.com/open?id=1IOu8GFxsn8uUyy96XoYpBVFcTIGmX6Op

I think that's not your latest attempt.
 
  • #18
PeroK said:
I think that's not your latest attempt.
Oh sorry.
This is the link: https://drive.google.com/open?id=1XHJIkiWteVRzxxOrPtf4g50eWOXOFCA8
 
  • #19
Julie Yum said:
Oh sorry.
This is the link: https://drive.google.com/open?id=1XHJIkiWteVRzxxOrPtf4g50eWOXOFCA8

Okay, so you have:

##v_y = u \sqrt{f^2 - \cos^2\theta} = u\sin\theta - \frac{gd}{u\cos\theta}##

Try multiplying thru by ##u## and you are nearly there.
 
  • #20
PeroK said:
Okay, so you have:

##v_y = u \sqrt{f^2 - \cos^2\theta} = u\sin\theta - \frac{gd}{u\cos\theta}##

Try multiplying thru by ##u## and you are nearly there.
Does this seem like a logical process?
https://drive.google.com/open?id=1c-cH_MS6Z1DKseQg6SRo2gYQE7W7nebw

And thanks for your help! It really helped me with getting through this question.
 
  • #21
Julie Yum said:
Does this seem like a logical process?
https://drive.google.com/open?id=1c-cH_MS6Z1DKseQg6SRo2gYQE7W7nebw

And thanks for your help! It really helped me with getting through this question.

You made a mistake there with ##-gd## becoming ##+gd##.

Also, it was a bit simpler

##u \sqrt{f^2 - \cos^2\theta} = u\sin\theta - \frac{gd}{u\cos\theta}##

##u(\sin\theta - \sqrt{f^2 - \cos^2\theta}) = \frac{gd}{u\cos\theta}##

##u^2(\sin\theta - \sqrt{f^2 - \cos^2\theta}) = \frac{gd}{\cos\theta}##

And that's good enough. You have numeric values for everything in that equation except ##u##.
 
  • #22
PeroK said:
You made a mistake there with ##-gd## becoming ##+gd##.

Also, it was a bit simpler

##u \sqrt{f^2 - \cos^2\theta} = u\sin\theta - \frac{gd}{u\cos\theta}##

##u(\sin\theta - \sqrt{f^2 - \cos^2\theta}) = \frac{gd}{u\cos\theta}##

##u^2(\sin\theta - \sqrt{f^2 - \cos^2\theta}) = \frac{gd}{\cos\theta}##

And that's good enough. You have numeric values for everything in that equation except ##u##.
Would I use -9.81 or +9.81 for g?
 
  • #23
Julie Yum said:
Would I use -9.81 or +9.81 for g?
How was g used in those kinematic equations?

In Post #6, @PeroK wrote:
PeroK said:
...

##v = fu##, hence ##v^2 = f^2u^2##

##v_x = u_x = u\cos\theta##,
##v_y = u_y - gt = u\sin\theta - gt##
...
That last equation Is like its one dimensional version: ##\ v = v_0 +at\,,\ ## where a is the acceleration.
So ##\ a=-g\,.\ ## Right?

So, what is acceleration? It's ##\ a=-9.81\,##m/s2.

Looks as though ##\ -g=-9.81\,##m/s2.
 

1. What is projectile motion?

Projectile motion is the motion of an object through the air or space under the influence of gravity. It is a special type of motion that combines both horizontal and vertical motion.

2. Can projectile motion be described with only distance, an angle, and a ratio?

Yes, projectile motion can be described with only distance, an angle, and a ratio. These three parameters are enough to determine the initial velocity and direction of the object, which can then be used to predict its motion.

3. How is the angle of projection related to the range of a projectile?

The angle of projection is the angle at which the object is launched into the air. The range of a projectile is the maximum horizontal distance it travels before hitting the ground. The range is directly proportional to the sine of the angle of projection, meaning that the greater the angle of projection, the greater the range.

4. What is the relationship between the initial velocity and the angle of projection in projectile motion?

The initial velocity and the angle of projection are directly related in projectile motion. The initial velocity can be broken down into horizontal and vertical components, with the horizontal component being determined by the cosine of the angle of projection and the vertical component determined by the sine of the angle of projection.

5. How does air resistance affect projectile motion?

Air resistance, also known as drag, can have a significant effect on projectile motion. It acts in the opposite direction of the object's motion and can cause the object to slow down and deviate from its predicted path. In most cases, air resistance is negligible, but in some cases, it must be taken into account to accurately describe the motion of a projectile.

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