Projectile motion with only distance, an angle and a ratio

[Julie Yum]

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

 

[stockzahn]

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?

 

[PeroK]

@Julie Yum this looks like quite a difficult problem. It will test your algebra and your mathematical understanding of the equations of projectile motion.

 

[Julie Yum]

@Julie Yum this looks like quite a difficult problem. It will test your algebra and your mathematical understanding of the equations of projectile motion.

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

 

[Julie Yum]

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.

 

[PeroK]

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?

 

[Julie Yum]

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?

 

[PeroK]

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.

 

[Julie Yum]

$u$ 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

 

[PeroK]

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.

 

[Julie Yum]

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

 

[PeroK]

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.

 

[Julie Yum]

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

 

[PeroK]

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.

 

[Ray Vickson]

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/

 

[Julie Yum]

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

 

[PeroK]

How would I continue to isolate for u?
https://drive.google.com/open?id=1IOu8GFxsn8uUyy96XoYpBVFcTIGmX6Op

I think that's not your latest attempt.

 

[Julie Yum]

I think that's not your latest attempt.

Oh sorry.
This is the link: https://drive.google.com/open?id=1XHJIkiWteVRzxxOrPtf4g50eWOXOFCA8

 

[PeroK]

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.

 

[Julie Yum]

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.

 

[PeroK]

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 $u$.

 

[Julie Yum]

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?

 

[SammyS]

Would I use -9.81 or +9.81 for g?

How was g used in those kinematic equations?

In Post #6, @PeroK wrote:

...

$v = fu$, 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.