Max initial angle so that the projectile range keeps increasing

[simphys]

Homework Statement

A projectile thrown from a point P moves in such a way
that its distance from P is always increasing. Find the maximum angle
above the horizontal with which the projectile could have been thrown.
Ignore air resistance.

Relevant Equations

projectile motion

Can I get a hint on how to start with this problem ? Thanks in advance
I can't really think of anything...

 

[ergospherical]

If there exists a point Q on the parabola where the distance from the launch point P is just about to start decreasing with time, the trajectory is orthogonal to the vector $\overrightarrow{PQ}$, yes? Which would make sense, because at this point the velocity has no radial component.

(To see this another way, if $d(r^2)/dt = d(x^2 + y^2)/dt = 2x \dot{x} + 2y\dot{y} = 0$ at this point, then $dy/dx = - x/y$, i.e. the gradient is orthogonal to that of the radius vector).

Does that help?

 

[PeroK]

Homework Statement:: A projectile thrown from a point P moves in such a way
that its distance from P is always increasing. Find the maximum angle
above the horizontal with which the projectile could have been thrown.
Ignore air resistance.
Relevant Equations:: projectile motion

Can I get a hint on how to start with this problem ? Thanks in advance
I can't really think of anything...

If you throw the object straight up, then it gets further away for a while, but then falls back towards you. That's no use.

If you throw the object almost along the ground, then it keeps getting further and never gets closer.

There must be some maximum angle in between the two where the object is always getting further away.

Use calculus to find it!

 

[simphys]

If there exists a point Q on the parabola where the distance from the launch point P is just about to start decreasing with time, the trajectory is orthogonal to the vector $\overrightarrow{PQ}$, yes? Which would make sense, because at this point the velocity has no radial component.

(To see this another way, if $d(r^2)/dt = d(x^2 + y^2)/dt = 2x \dot{x} + 2y\dot{y} = 0$ at this point, then $dy/dx = - x/y$, i.e. the gradient is orthogonal to that of the radius vector).

Does that help?

If you throw the object straight up, then it gets further away for a while, but then falls back towards you. That's no use.

If you throw the object almost along the ground, then it keeps getting further and never gets closer.

There must be some maximum angle in between the two where the object is always getting further away.

Use calculus to find it!

owh goshhh! Thanks a lot guys. So I kept looking at it but didn't really get why using that distance r^2 until it hit me that it's the distance/(magnitude of the position vector) that need not to become negative i.e. derivate >= 0... and not one of the components or something like that. I was trying to get some relation going via the components to find that angle lol.

One question though. Can you please (@ergospherical) explain that sentence where you said that the trajectory is orthogonal to the vector a bit more please? Why is that? I will show you what I think you mean but I don't get why that's the case.

Thanks a lot oncde again

 

[simphys]

This

 

[PeroK]

owh goshhh! Thanks a lot guys. So I kept looking at it but didn't really get why using that distance r^2 until it hit me that it's the distance/(magnitude of the position vector) that need not to become negative i.e. derivate >= 0... and not one of the components or something like that. I was trying to get some relation going via the components to find that angle lol.

One question though. Can you please (@ergospherical) explain that sentence where you said that the trajectory is orthogonal to the vector a bit more please? Why is that? I will show you what I think you mean but I don't get why that's the case.

Thanks a lot oncde again

I don't want to put you off, but this problem might be a little bit advanced. There is a direct approach, which is just to crunch the calculus. The other approach involves first finding an equation for the parabola $y(x)$ by eliminating $t$. If you haven't seen that already.

 

[ergospherical]

This

Good, yes!

Do you recall the "SUVAT" equations of constant acceleration? Can you write the velocity and position as functions of time, i.e. $\mathbf{v}(t) = (v_x(t), v_y(t))$ and $\mathbf{r}(t) = (x(t),y(t))$. Once you have accomplished this you can put $\mathbf{r} \cdot \mathbf{v} = 0$ and employ what you know about the discriminants of quadratic equations (so as to enforce that there are no solutions...).

 

[simphys]

I don't want to put you off, but this problem might be a little bit advanced. There is a direct approach, which is just to crunch the calculus. The other approach involves first finding an equation for the parabola $y(x)$ by eliminating $t$. If you haven't seen that already.

it is indeed a challange problem haha, last 3 exercises of the chapters are challenge problems.
(and you don't put me off in any way! :) that's why I am here to learn haha)

So I am guessing the direct approach is looking at the derivate of the that distance r.

And well I have used this relationship, but not in this book(I remember it was I think specifically for using y(x) and then getting a relation for dy/dt and dx/dt or the velocities in both directions and getting them in such a way) , so I kinda know what you mean with this relationship --> find y in terms of x = losing the parameters and but then, what indicator should I look for afterwards? thanks

 

[simphys]

Good, yes!

Do you recall the "SUVAT" equations of constant acceleration? Can you write the velocity and position as functions of time, i.e. $\mathbf{v}(t) = (v_x(t), v_y(t))$ and $\mathbf{r}(t) = (x(t),y(t))$. Once you have accomplished this you can put $\mathbf{r} \cdot \mathbf{v} = 0$ and employ what you know about the discriminants of quadratic equations (so as to enforce that there are no solutions...).

hoooww that is so amazing! didn't think about the scalar product at all...
But @ergospherical can you please explain why we know that'd the be the case? that the vector is orthogonal to the trajectory?

Edit: Thank you, I indeed found the relation with answer 70.5 degrees. appreciate it a lot!

 

[simphys]

So If I'd look at it in terms of the fact that the derivate of the distance can become zero as perok pointed that's easy

But for this method, how we actually know that v and position are orthogonal is a question for me at this point

 

[ergospherical]

hoooww that is so amazing! didn't think about the scalar product at all...
But @ergospherical can you please explain why we know that'd the be the case? that the vector is orthogonal to the trajectory?

Edit: Thank you, I indeed found the relation with answer 70.5 degrees. appreciate it a lot!

Lovely stuff.

There's a few ways to look at it. The rate of change of distance is the radial component of the velocity; if $\mathbf{v}$ is orthogonal to $\mathbf{r}$, then $\mathbf{v}$ has no radial component.

You could also think in terms of calculus;\begin{align*}
\dfrac{d(r^2)}{dt} = \frac{d}{dt} (\mathbf{r} \cdot \mathbf{r}) = 2\mathbf{r} \cdot \mathbf{v}
\end{align*}so turning points of $r^2$ (and hence $r$) correspond to when $\mathbf{r} \cdot \mathbf{v} = 0$.

 

[simphys]

Lovely stuff.

There's a few ways to look at it. The rate of change of distance is the radial component of the velocity; if $\mathbf{v}$ is orthogonal to $\mathbf{r}$, then $\mathbf{v}$ has no radial component.

You could also think in terms of vectors;\begin{align*}
\dfrac{d(r^2)}{dt} = \frac{d}{dt} (\mathbf{r} \cdot \mathbf{r}) = 2\mathbf{r} \cdot \mathbf{v}
\end{align*}so turning points of $r^2$ (and hence $r$) correspond to when $\mathbf{r} \cdot \mathbf{v} = 0$.

noo way... That is epic. I haven't learnd that one yet wow.. Thanks a lot! I appreciate the explanation as well!
That is interesting so you can divide not only the acceleration but also the velocity in n-t components (normal/tangential)

 

[ergospherical]

You are most welcome. Well done on solving the problem so quickly!

 

[simphys]

You are most welcome. Well done on solving the problem so quickly!

Thank you, I can now proceed with forces once again hehe.

By the way, (as this is introductory physics) when should I actually complement this with a classical mechanics book or should I better just stick with this one? I am at university so I would just go through it in the left free time.

I won't be doing rotational motion and dynamics or rigit bodies at this point because I don't have the time to at the moment. I'll only do dynamics of particles and kinetics and energy and temperature and heat as that's what needs to be known for electricity and stuff like that starting from the 2nd semester. (to keep in mind this is my first physics intro)

 

[ergospherical]

It all depends on your goals. In my first year we were given a comprehensive set of lecture notes for the course, but I studied some extra (non-assessable) material from D. Gregory's text and L. Landau's mechanics text to learn how to solve problems in general rigid body mechanics and analytical mechanics. I found that getting this head-start put me in a favourable position when we later came to study these topics formally the next year.

If in doubt, discuss with your supervisors. It never hurts to read ahead, though - especially in Physics! Borrow a selection of mechanics books from the university library and see if you warm to any of them.

 

[simphys]

It all depends on your goals. In my first year we were given a comprehensive set of lecture notes for the course, but I studied some extra (non-assessable) material from D. Gregory's text and L. Landau's mechanics text to learn how to solve problems in general rigid body mechanics and analytical mechanics. I found that getting this head-start put me in a favourable position when we later came to study these topics formally the next year.

If in doubt, discuss with your supervisors. It never hurts to read ahead, though - especially in Physics! Borrow a selection of mechanics books from the university library and see if you warm to any of them.

Thanks a lot! Will definitely do. I guess that rigit bodies will be lots of geometry as well, so once I finish that I'll get into that part for sure.

 

[simphys]

@ergospherical
As a matter of relevance,
I was just googling for a bit about a guide/ curriculum on physics,
Found this one
That honestly might not be a bad one to follow on the side, what do you think

 

[ergospherical]

There's some nice books there, but the usual problem is that when you ask 10 different people what the best textbook on a certain topic is it's not unlikely that you'll get 10 different responses. It's quite hard to go wrong - my best advice would be to go down to the library, check out a dozen-or-so books and see what clicks for you! If you're stuck for ideas, check out the textbook recommendation section of the forum (where there's an abundance of such threads).

 

[simphys]

There's some nice books there, but the usual problem is that when you ask 10 different people what the best textbook on a certain topic is it's not unlikely that you'll get 10 different responses. It's quite hard to go wrong - my best advice would be to go down to the library, check out a dozen-or-so books and see what clicks for you! If you're stuck for ideas, check out the textbook recommendation section of the forum (where there's an abundance of such threads).

facts! Thank you will definitely do that. Probably look up some recommendations online first and look and compare afterwards which I would deem best to study, at the end of the day the book is not the most most important, but the method of how you study is have I found.