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

[neilparker62]

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.

 

[malawi_glenn]

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.

 

[hutchphd]

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...

 

[malawi_glenn]

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?

 

[hutchphd]

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.

 

[malawi_glenn]

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.

 

[kuruman]

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.

 

[neilparker62]

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 $$\tan\theta = \frac{|u|}{|v|}$$

 

[kuruman]

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.

 

[Orodruin]

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.

 

[neilparker62]

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.

 

[neilparker62]

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!