# Initial velocity and angle when a ball is kicked over a 3m fence

• simphys
In summary: The answer seems correct based on what is shown in the video. However, it's always good to check things yourself.
erobz said:
Just so I have some better understanding of policy, have I subtly crossed the line? My thought was if they start thinking about this problem (regardless of if they succeed in solving it or not) it would help them understand the HW problem.
It's just that it is much better to keep new questions/problems in their own thread. It's too confusing for helpers if threads change subject mid-thread. • erobz
Consider that
$$x = vt\cos\theta, \qquad y = vt \sin\theta -\frac{gt^2}{2}.$$
Substituting the first into the second
$$y = x \tan\theta - \frac{gx^2}{2v^2\cos^2\theta}.$$
For a fixed ##x## and fixed ##v##, this is the height ##y## at position ##x## when launched at angle ##\theta## with speed ##v##. To find the equation of the envelope, we need to maximise this wrt ##\theta##.
$$\frac{dy}{d\theta} = \frac{x}{\cos^2\theta} - \frac{gx^2\sin\theta}{v^2\cos^3\theta} = x \frac{v^2 \cos\theta - gx \sin\theta}{v^2 \cos^3\theta}= 0.$$
Implying
$$\tan\theta = \frac{v^2}{gx}, \quad \frac{1}{\cos^2\theta} = 1 + \tan^2\theta = 1 + \frac{v^4}{g^2 x^2}$$
and therefore
$$y_{\rm max} = \frac{v^2}{g} - \frac{gx^2}{2v^2}\left( 1 + \frac{v^4}{g^2 x^2}\right) = \frac{v^2}{2g} - \frac{gx^2}{2v^2}.$$

So, why is this interesting to the problem at hand? Well, since we now have the highest you can reach at ##x## with speed ##v##, we just need to find the speed ##v## such that ##y_{\rm max} = \Delta y## when ##x = \Delta x##. This is a straightforward second order polynomial equation in ##v^2##:
$$(v^2)^2 - 2g \Delta y \, v^2 - g^2 \Delta x^2 = 0$$
and therefore
$$v^2 = g \Delta y + g\sqrt{\Delta y^2 + \Delta x^2}.$$

… assuming I managed to avoid arithmetic errors while doing this on my phone …

Edit: Note that the angle ##\theta## as found above of course councides with the expression given by @kuruman once the solution for ##v^2## is inserted into the expression for ##\tan\theta##.

I’d therefore say that the solution to the question posed in #33 is not only a curiosity related to the original problem but an actual viable way to solve it.

• hutchphd and erobz
I agree with your assessment and the equation in the spoiler. There are no arithmetic errors. It is the optimization condition, equation (6), that I derived in my insight article. I did not show it here because I assumed that this is still a "live" homework problem.

If two of the three variables are given, solving this condition for the third will give the optimized value, maxima for the displacements and minimum for the speed. For example, it's a one-liner solution to the oft seen homework problem of maximizing the range of a projectile fired off a cliff when the initial speed and cliff height are given.

• hutchphd and erobz
Orodruin said:
… assuming I managed to avoid arithmetic errors while doing this on my phone …

kuruman said:
I agree with your assessment and the equation in the spoiler. There are no arithmetic errors. It is the optimization condition, equation (6), that I derived in my insight article.
You didn't seem to use calculus though to get the optimization condition. Correct?

Correct.

erobz said:
No, I did it in LaTeX code in a text box on an iPhone.

• erobz
erobz said:
Physics, it's just in our heads :(

• erobz
kuruman said:
$$\theta=\arctan\left[\sqrt{1+\left(\frac{\Delta y}{\Delta x}\right)^2}+\frac{\Delta y}{\Delta x}\right].$$
If anyone's interested, here's an alternative - derived using the method in the video in @simphys's Post #12. It agrees with @kuruman's above formula (checked numerically).$$2\theta = \pi - \arctan \left( \frac {\Delta x}{\Delta y} \right)$$

• PeroK
simphys said:
Homework Statement:: Determine the minimum initial velocity v0 at which the ball must be kicked in order for it to just cross over the 3-m high fence .

Part B:
Determine the corresponding angle ##\theta _0## at which the ball must be kicked.
Use following vector product equation: $$\vec{a} \times \vec{s} = \vec{v} \times \vec{u}$$ with:

##\vec{a}=(0,-g)##
##\vec{s}=(6,3)##
##\vec{u}=(u\cos(\theta), u\sin(\theta))##
##\vec{v}=(\sqrt{(u^2-6g)}\sin(\theta), -\sqrt{(u^2-6g)}\cos(\theta))##
and then $$\tan\theta=\frac{u}{v}$$. Determine u and hence v from the vector equation. Do NOT attempt to substitute for ##\theta## when processing the vector equation.

The caveat here is that probably no-one will know what you're talking about! But in essence we are using the method mentioned by @kuruman in post #19:

the velocity at the point of interest (here when passing over the wall) will be perpendicular to the initial velocity Last edited:
simphys said:
Thank you, that is interesting. I don't understand how you think in such a way by taking ##-x_0## and stuff, I would never ever do such a thing xD

And thanks for the conclusion, you are right.. good attitude to have to my own answers as well tbh.
But yeah, the thing is that this exercise has probably been done by 600 + students so I kinda assume that all of the answers should be right, no? The answers you quoted are indeed correct.

Steve4Physics said:
If anyone's interested, here's an alternative - derived using the method in the video in @simphys's Post #12. It agrees with @kuruman's above formula (checked numerically).$$2\theta = \pi - \arctan \left( \frac {\Delta x}{\Delta y} \right)$$
We can do even better. We let ##\alpha## be the direct angle to the target point. I.e. ##\tan \alpha = \frac{\Delta y}{\Delta x}##. Note that ##\frac {\Delta x}{\Delta y} = \tan(\frac \pi 2 - \alpha)##. Hence:
$$\theta = \frac \pi 4 + \frac \alpha 2 \ \ \ \ (0 \le \alpha \le \frac \pi 2)$$Moreover, we have:
$$\tan \theta = \tan \alpha + \sec \alpha$$$$v^2 = g\Delta x \tan \theta = g\Delta x(\tan \alpha + \sec \alpha)$$

• neilparker62 and Steve4Physics
PeroK said:
We can do even better. We let ##\alpha## be the direct angle to the target point. I.e. ##\tan \alpha = \frac{\Delta y}{\Delta x}##. Note that ##\frac {\Delta x}{\Delta y} = \tan(\frac \pi 2 - \alpha)##. Hence:
$$\theta = \frac \pi 4 + \frac \alpha 2 \ \ \ \ (0 \le \alpha \le \frac \pi 2)$$Moreover, we have:
$$\tan \theta = \tan \alpha + \sec \alpha$$$$v^2 = g\Delta x \tan \theta = g\Delta x(\tan \alpha + \sec \alpha)$$
Indeed. The last equation is equation is equation (6) in my insight with ##\alpha## replacing ##\varphi## and a bit of rearrangement. Another useful expression relating the three relevant angles to projectile motion,
##\theta =~##angle of projection
##\varphi =~##direct angle to the target point
##\omega =~##angle of velocity relative to the horizontal at the target point
is $$\tan\varphi=\frac{1}{2}(\tan\theta+\tan\omega).$$ This is easy to remember as "the tangent of the final position vector angle is the average of the tangents of the initial velocity and the velocity-at-target angles."

The trajectory is optimized when the initial velocity and the velocity at target are perpendicular, ##(\theta+\omega=\pi/2)## which results in the relation between ##\theta## and ##\varphi## at optimum. It's a pity that the canonical method for solving projectile problems is to start with the two kinematic equations, find the time to the target and use that to find anything else one is interested in. The canonical method is guaranteed to work, but a lot can be done more efficiently and cleanly if one introduces the direct angle to the target point in the equations. I discuss all that in my insight and provide illustrations of the use of these ideas to solve problems drawn from several threads.

• hutchphd and PeroK
@kuruman that's a great Insight article!

• hutchphd and kuruman
Thanks. I wrote it hoping to provide an alternative to the canonical way of skinning the proverbial cat. I hope it catches on.

PeroK said:
We can do even better. We let ##\alpha## be the direct angle to the target point. I.e. ##\tan \alpha = \frac{\Delta y}{\Delta x}##. Note that ##\frac {\Delta x}{\Delta y} = \tan(\frac \pi 2 - \alpha)##. Hence:
$$\theta = \frac \pi 4 + \frac \alpha 2 \ \ \ \ (0 \le \alpha \le \frac \pi 2)$$Moreover, we have:
$$\tan \theta = \tan \alpha + \sec \alpha$$$$v^2 = g\Delta x \tan \theta = g\Delta x(\tan \alpha + \sec \alpha)$$
Find ##u^2## if ##Rg=u.\sqrt{u^2-2gh}##.

Last solution applicable. Times through by ##\frac{R}{R}## to rework into the form given by @PeroK above.

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malawi_glenn said:
I think it also has to do with the actual "passing over" phrase. When presented like they were just going to hit a target 6.0 m away from them at a 3.0 m height, the success rate would be higher.
@hutchphd I actually teach two different classes next academic year. They are about the same level. I might give them this problem on their test (otherwise identical tests) and will change the phrasing as I mentioned. Could be interesting.

Yeah maybe that would be better. Is there any other way to lead them down the correct path? I think this will be tough on a test...

hutchphd said:
Yeah maybe that would be better. Is there any other way to lead them down the correct path? I think this will be tough on a test...
You think the problem in this thread is too hard to be put on a test?

It would depend upon the class.! I always enjoyed having (at least part of) a problem that I only expected a few folks to get.

hutchphd said:
It would depend upon the class.! I always enjoyed having (at least part of) a problem that I only expected a few folks to get.
I have very good classes. I usually give 2/3 of the students highest grade in intro physics. On my former school I would give 2/3 a passing grade or higher (and the rest fail).
So I think this kind of problem will not be too diffuclt for them, and I can compare if the problem situation would make them more inclined to think wrong.

• hutchphd
malawi_glenn said:
You think the problem in this thread is too hard to be put on a test?
You could tell what percentage of your students is in the right frame of mind if you ask a conceptual question based on the problem. I suggest that you show them the correct trajectory (orange line) as a question.

Question: Both projectiles just barely go over the wall. The projectile in the blue trajectory is traveling parallel to the ground when it does so. Is it possible that the initial speed of the projectile in the orange trajectory is less than that of the projectile in the blue trajectory? Explain your reasoning.

It's a subtle point that needs to be understood before tackling this problem. • PeroK and hutchphd
kuruman said:
The correct angle is not 45°, it is $$\theta=\arctan\left[\sqrt{1+\left(\frac{\Delta y}{\Delta x}\right)^2}+\frac{\Delta y}{\Delta x}\right].$$
Or: $$\theta=\arctan \sqrt{\frac{d+h}{d-h}}$$ where d is the direct distance to the top of the fence from the launch point ##(=\sqrt {\Delta x^2 + \Delta y^2})## and ##h=\Delta y.## Also: $$|u|=\sqrt{g(d+h)}$$ $$|v|=\sqrt{g(d-h)}$$ - consistent with $$\tan\theta = \frac{|u|}{|v|}$$

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• kuruman
neilparker62 said:
Or: $$\theta=\arctan \sqrt{\frac{d+h}{d-h}}$$ where d is the direct distance to the top of the fence from the launch point ##(=\sqrt {\Delta x^2 + \Delta y^2})## and ##h=\Delta y.## Also: $$|u|=\sqrt{g(d+h)}$$ $$|v|=\sqrt{g(d-h)}$$ - consistent with $$\tan\theta = \frac{|u|}{|v|}$$
Interesting. We know that at trajectory optimization ##\mathbf{u}## and ##\mathbf{v}## are perpendicular. It follows that $$u^2+v^2=d(d+h)+g(d-h)=2gd\tag{1}$$ We also know that the two perpendicular vectors are related by the kinematic equation $$\mathbf{v}-\mathbf{u}=\mathbf{g}~t \implies v^2+u^2=g^2t_{\!f}^2\tag{2}$$ Putting the two equations together gives the time of flight at optimization, $$t_{\!f}=\sqrt{\frac{2d}{g}}$$ where d is the length of the displacement vector.

This completes the condition for optimizing one of ##\Delta x##, ##\Delta y## or ##v_0## as the projection angle is varied:

"At optimization,(a) the projection velocity and the velocity at target are perpendicular and (b) the transit time is the same as if the projectile were dropped from rest a distance equal to the magnitude of the overall displacement vector."

I am amazed that there is still juice left to squeeze out of projectile motion.

kuruman said:
I am amazed that there is still juice left to squeeze out of projectile motion.
My Master thesis supervisor - who would have been my PhD supervisor if he had not been close to retirement - used to say that you can pick any university level physics textbook, randomly open a page and point to a formula and write a paper about it or some details stemming from it.

kuruman said:
This completes the condition for optimizing one of ##\Delta x##, ##\Delta y## or ##v_0## as the projection angle is varied:

"At optimization,(a) the projection velocity and the velocity at target are perpendicular and (b) the transit time is the same as if the projectile were dropped from rest a distance equal to the magnitude of the overall displacement vector."

I am amazed that there is still juice left to squeeze out of projectile motion. See also Winans Equations 20 - 27.

• kuruman
Instead of being amazed I should have reread Winans. Nevertheless, I think it's good to bring forth this simple statement of trajectory optimization. It is more likely to be noticed here than in a 1961 AJP article.

kuruman said:
Instead of being amazed I should have reread Winans. Nevertheless, I think it's good to bring forth this simple statement of trajectory optimization. It is more likely to be noticed here than in a 1961 AJP article.
It most definitely is. Not to mention ##\vec{a}\times \vec{s} = \vec{v} \times \vec{u}## which I 'punt' whenever I can!

• kuruman