Rock touching wall

[compaq65]

Homework Statement:

Rock is thrown at an angle and speed so that he almost touches three walls. Distance between walls are r and 2r (from left). Wall in the middle is two times higher than two other equal walls. Rock's flight range is nr. Find n.

Relevant Equations:

kinematic eqs.

I tried to write a trajectory equation of motion.


$$x(t)=vcos\alpha t$$

$$y(t)=vsin\alpha t-\frac{1}{2}gt^{2}$$


from these we get:


$$y=xtan\alpha -\frac{x^{2}g(1+tan^{2}\alpha )}{2v^{2}}$$


For this problem:


$$h=xtan\alpha -\frac{x^{2}g(1+tan^{2}\alpha )}{2v^{2}}$$


$$2h=(x+r)tan\alpha -\frac{(x+r)^{2}g(1+tan^{2}\alpha )}{2v^{2}}$$


$$h=(x+3r)tan\alpha -\frac{(x+3r)^{2}g(1+tan^{2}\alpha )}{2v^{2}}$$


$$0=nrtan\alpha-\frac{(nr)^{2}g(1+tan^{2}\alpha )}{2v^{2}} \Rightarrow tan\alpha= \frac{nrg(1+tan^{2}\alpha )}{2v^{2}}$$


So, we have 4 equations and 5 unknowns (x, tanα, v, h, n) and I stuck at solving them. May somebody can help?

I have an idea of adding $$nr=\frac{v^{2}sin2\alpha }{g}.$$ But then I surely don't know how to solve it.

Also, can we approximate that $$tan\alpha =\frac{h}{x} ?$$ I believe that would help solving this problem.

 

[PeroK]

What about looking at the middle section of the trajectory, taking the top of the first wall as the origin of the trajectory? That might give you a relationship between $r$ and $h$.

 

[TSny]

I have an idea of adding $$nr=\frac{v^{2}sin2\alpha }{g}.$$

This equation is equivalent to your 4^th equation $\tan\alpha= \frac{nrg(1+\tan^{2}\alpha )}{2v^{2}}$

 

[TSny]

From the diagram, you can see how $nr$ is related to $r$ and $x$.

 

[compaq65]

From the diagram, you can see how $nr$ is related to $r$ and $x$.

$$n=\frac{2x}{3}+3?$$
Ok, but still missing one equation somewhere...

 

[PeroK]

$$n=\frac{2x}{3}+3?$$
Ok, but still missing one equation somewhere...

That's $$n=\frac{2x}{r}+3$$

I still think it's easier to analyse the middle section.

 

[compaq65]

That's $$n=\frac{2x}{r}+3$$

I still think it's easier to analyse the middle section.

I tried, but we still don't know the highest point, angle or speed. Though it is good that we know that horizontal distance is 3r, I think trajectory equation wouldn't help here. Or should I use any other equations?

 

[PeroK]

I tried, but we still don't know the highest point, angle or speed. Though it is good that we know that horizontal distance is 3r, I think trajectory equation wouldn't help here. Or should I use any other equations?

I think your $x$ is a) unnecessary and b) clashes with using $x$ as the general horizontal coordinate. Also, I'd take the "initial" speed, $v$, and angle, $\theta$, at the top of the first wall and make that $x=0, y = 0$. That gives you:
$$y = x\tan \theta - \frac{gx^2}{2v^2}\sec^2 \theta$$That's valid for the whole trajectory, by the way. Then, we know that the range for that middle section is $3r$. That should allow you to eliminate $v^2$

Then, we know that at $x = r$ we have $y = h$. That should allow you to eliminate $\tan \theta$.

 

[pbuk]

I still think it's easier to analyse the middle section.

So do I.

Or should I use any other equations?

What shape is the curve? Do you know any equations for that shape?

Spoiler

If you measure coordinates from an origin at height $h$ up the middle wall the top of one of the walls it makes the maths easier.

 

[compaq65]

I think your $x$ is a) unnecessary and b) clashes with using $x$ as the general horizontal coordinate. Also, I'd take the "initial" speed, $v$, and angle, $\theta$, at the top of the first wall and make that $x=0, y = 0$. That gives you:
$$y = x\tan \theta - \frac{gx^2}{2v^2}\sec^2 \theta$$That's valid for the whole trajectory, by the way. Then, we know that the range for that middle section is $3r$. That should allow you to eliminate $v^2$

Then, we know that at $x = r$ we have $y = h$. That should allow you to eliminate $\tan \theta$.

Ok, after eliminating $v^2$ and after simplifying $$sin2\theta sec^{2}\theta\Rightarrow 2tan\theta$$
I get $$h=rtan\theta (1-\frac{1}{3g}).$$
Didn't really understand how I can eliminate $\tan \theta$.

 

[PeroK]

Ok, after eliminating $v^2$ and after simplifying $$sin2\theta sec^{2}\theta\Rightarrow 2tan\theta$$
I get $$h=rtan\theta (1-\frac{1}{3g}).$$

That's dimensionally inconsistent. You can't have $1-\frac{1}{3g}$.

Didn't really understand how I can eliminate $\tan \theta$.

That equation (when corrected) allows you to replace $\tan \theta$ in your parabolic equation.

 

[jbriggs444]

Personally, I always opt to extract as much from geometry or symmetry as possible before shifting to algebra. The horizontal position of the apex of the shape should be obvious. With that in hand, there are further relationships that can be inferred.

 

[compaq65]

That's dimensionally inconsistent. You can't have $1-\frac{1}{3g}$.

That equation (when corrected) allows you to replace $\tan \theta$ in your parabolic equation.

Sorry, $$tan\theta =\frac{6h}{5r}.$$

But parabolic equation involves velocity, isn't it?

 

[Steve4Physics]

Can I add (a little more explicitly) to what others have already hinted?

The trajectory is a parabola. You know the positions of 3 points through which the parabola passes, so the equation of the parabola is fully determined.

(It makes sense to try to choose a suitable point as the origin in order to make the equation as simple as possible.)

The problem can then be solved without any angles, speeds or trig' functions.

 

[PeroK]

Sorry, $$tan\theta =\frac{6h}{5r}.$$

That's not what I get.

But parabolic equation involves velocity, isn't it?

You can use the range formula, for example, to eliminate $v^2$. Which I thought you'd done.

 

[compaq65]

That's not what I get.

You can use the range formula, for example, to eliminate $v^2$. Which I thought you'd done.

I wrote 2 equations for that middle section of trajectory:
$$h=rtan\theta -\frac{gr^{2}sec^{2}\theta }{2v^{2}}\hspace{4 mm}(1)$$
$$0=3rtan\theta -\frac{9gr^{2}sec^{2}\theta }{2v^{2}}\hspace{4 mm}(2)$$
from these:

$$2v^{2}=\frac{3grsec^{2}\theta }{tan\theta }$$

$$2v^{2}=\frac{gr^{2}sec^{2}\theta }{rtan\theta-h }$$

eliminating $2v^2$

$$\frac{3}{tan\theta }=\frac{r}{rtan\theta -h}\hspace{4mm}(3)$$
$$tan\theta =\frac{3h}{2r}$$

So I'm stuck here. I have tangent, but I can't put it back in $(3)$.

And I can't use range formula
$$3r=\frac{v^{2}sin2\theta }{g}$$
because it is same as $(2)$.

I suppose I also can't write any more equations in this part of trajectory apart those, when $y=0$ and $y=h$. What I'm missing?

 

[PeroK]

I wrote 2 equations for that middle section of trajectory:
$$h=rtan\theta -\frac{gr^{2}sec^{2}\theta }{2v^{2}}\hspace{4 mm}(1)$$
$$0=3rtan\theta -\frac{9gr^{2}sec^{2}\theta }{2v^{2}}\hspace{4 mm}(2)$$
from these:

$$2v^{2}=\frac{3grsec^{2}\theta }{tan\theta }$$

$$2v^{2}=\frac{gr^{2}sec^{2}\theta }{rtan\theta-h }$$

eliminating $2v^2$

$$\frac{3}{tan\theta }=\frac{r}{rtan\theta -h}\hspace{4mm}(3)$$
$$tan\theta =\frac{3h}{2r}$$

That should allow you to write the equation for $y$ in terms of $x, h$ and $r$. Eliminating both $v$ and $\theta$.

I suppose I also can't write any more equations in this part of trajectory apart those, when $y=0$ and $y=h$. What I'm missing?

That equation holds for the entire parabola, including outside the middle section. The "initial" velocity and angle can be chosen at any point in the trajectory.

So, you can also solve for $y = -h$ and obtain the two values of $x$ where the object starts and ends.

 

[compaq65]

That should allow you to write the equation for $y$ in terms of $x, h$ and $r$. Eliminating both $v$ and $\theta$.

So you mean, that if I know tanθ, then I somehow can rewrite
$$y = x\tan \theta - \frac{gx^2}{2v^2}\sec^2 \theta$$
in terms of y, x, h, r?

 

[PeroK]

So you mean, that if I know tanθ, then I somehow can rewrite
$$y = x\tan \theta - \frac{gx^2}{2v^2}\sec^2 \theta$$
in terms of y, x, h, r?

Precisely. When we saw "know" $\tan \theta$ that means express it in terms of $r$ and $h$. Which you have done.

 

[compaq65]

Precisely. When we saw "know" $\tan \theta$ that means express it in terms of $r$ and $h$. Which you have done.

Something like that?
$$v^{2}=\frac{3rg}{sin2\theta }$$
$$y=\frac{3hx}{2r}-\frac{gx^{2}sec^{2}\theta sin2\theta }{6rg}$$
$$y=\frac{3hx}{2r}-\frac{x^{2}tan\theta }{3r}$$
$$y=\frac{3hx}{2r}-\frac{x^{2}h}{2r^2}$$

 

[PeroK]

$$y=\frac{3hx}{2r}-\frac{x^{2}h}{2r^2}$$

You got there at last. Just one more step now.

 

[pbuk]

You got there at last.

The hard way, by far. Easier to use the three points marked below to solve for the constants in the parabolic equation $y = ax^2 + bx + c$

Point B for instance immediately gives $2h = (0^2)ax + 0bx + c \implies c = 2h$ (I am using $x$ here as the horizontal axis instead of the small distances at the ends). The problem is done in another half-dozen lines of similarly trivial algebra.

 

[PeroK]

The hard way, by far. Easier to use the three points marked below to solve for the constants in the parabolic equation $y = ax^2 + bx + c$
View attachment 319122
Point B for instance immediately gives $2h = (0^2)ax + 0bx + c \implies c = 2h$ (I am using $x$ here as the horizontal axis instead of the small distances at the ends). The problem is done in another half-dozen lines of similarly trivial algebra.

It's true that since we didn't need $v$ or $\theta$, then using generic coefficients ($a, b, c$) is simpler. The algebraic steps themselves, however, are the same.

 

[compaq65]

Can I find travel range by adding distances which are to the left and to the right of first wall. setting top of the first wall as $(0;0)$?

$$-h=\frac{3hx_{1}}{2r}-\frac{x_{1}^{2}h}{2r^{2}}$$
$$x_{1}^{2}-3rx_{1}-2r^{2}=0$$
$$x_{1}=\frac{r(3-\sqrt{17})}{2}\hspace{4 mm} and\hspace{4 mm}x_{2}=\frac{r(3+\sqrt{17})}{2}$$
$$x_{2}-x_{1}=r\sqrt{17}$$
$$nr=r\sqrt{17}$$
$$n=\sqrt{17}$$

 

[compaq65]

The hard way, by far. Easier to use the three points marked below to solve for the constants in the parabolic equation $y = ax^2 + bx + c$
View attachment 319122
Point B for instance immediately gives $2h = (0^2)ax + 0bx + c \implies c = 2h$ (I am using $x$ here as the horizontal axis instead of the small distances at the ends). The problem is done in another half-dozen lines of similarly trivial algebra.

What coefficients exactly mean in this specific problem? How can I solve them?

 

[PeroK]

What coefficients exactly mean in this specific problem? How can I solve them?

The idea is that it doesn't matter that the coefficients of the parabola can be written interms of $v^2$ and $\tan \theta$. We didn't need those. Using the same origin we can simply write:
$$y = x\tan \theta - \frac{gx^2}{2v^2}\sec^2 \theta = bx - ax^2$$Then solve for $a$ and $b$. This makes the algebra considerably easier to manage.

 

[erobz]

From the diagram it appears that the 2 zeros of the parabolic function are: $0, nr$.

So, you can surmise that:

$$ y(x) = a x ( x- nr) = a x^2 - anr x$$

Then you can solve for $x(r)$ from looking at total range:

$$2x + 3r = nr \implies x = \frac{nr - 3r}{2}$$

Then you solve the following system:

$$ \begin{align} h &= a \left( \frac{nr - 3r}{2}\right)^2 - anr \left( \frac{nr - 3r}{2}\right) \tag*{} \\ 2h &= a\left( \frac{nr - 3r}{2} + r \right)^2 - anr \left( \frac{nr - 3r}{2} + r\right) \tag*{} \end{align} $$

I think that just reduces to solving ( an ugly ) quadratic in $n$ in terms of $r$.

EDIT: the $a$ and the $r$ all get factored out in the simplification.

 

[compaq65]

The hard way, by far. Easier to use the three points marked below to solve for the constants in the parabolic equation $y = ax^2 + bx + c$

Point B for instance immediately gives $2h = (0^2)ax + 0bx + c \implies c = 2h$ (I am using $x$ here as the horizontal axis instead of the small distances at the ends). The problem is done in another half-dozen lines of similarly trivial algebra.

Why there is a $c$ constant, if in
$$y = x\tan \theta - \frac{gx^2}{2v^2}\sec^2 \theta$$
I can only notice $a$ and $b$ coefficients. Because of that I get a little bit different answers. Which method is correct?

 

[pbuk]

I can only notice $a$ and $b$ coefficients. Because of that I get a little bit different answers. Which method is correct?

There are many different equations that can describe the path, each with a different origin. In mine the coefficients are probably simpler but you need to work out $x$ (the roots) separately, in the one you are following you get it all at once.

 

[PeroK]

Why there is a $c$ constant, if in
$$y = x\tan \theta - \frac{gx^2}{2v^2}\sec^2 \theta$$
I can only notice $a$ and $b$ coefficients. Because of that I get a little bit different answers. Which method is correct?

@pbuk took a different origin, where at $x = 0$, $y = 2h$. So, $c = 2h$. It was even simpler to have the origin on the parabola itself at the top of the first wall, so that $c = 0$

 

[pbuk]

It was even simpler to have the origin on the parabola itself at the top of the first wall, so that $c = 0$

Yes that is a better choice.

 

[nasu]

Why there is a $c$ constant, if in
$$y = x\tan \theta - \frac{gx^2}{2v^2}\sec^2 \theta$$
I can only notice $a$ and $b$ coefficients. Because of that I get a little bit different answers. Which method is correct?

You should realize by now that this problem has nothing to do with the kinematics of the projectile. Velocity and acceleration are irrelevant. Your answer does not depend on them. The same question can be asked about a parabola painted on a wall.

 

[compaq65]

Everything is clear now. Thanks, you all helped me a lot!