Kinematic Problem w/ Parabola: Solving w/ KE Theorem?

[h1a8]

Homework Statement

A ball is thrown at an upward angle from an initial height 'h' above the ground. When the ball reaches a horizontal distance 'A' from the release point, it is at a maximum height 'H' above the ground. The ball eventually strikes the ground a horizontal distance 'R' from the release point. Assuming no drag, a horizontal ground, and a constant downward force of gravity (unknown), what is the height 'h' in which the ball was thrown in terms of 'H', 'A', and 'R'?

Relevant Equations

y_f=y_0+v_0 sinθt- 1/2 a_y t^2
x_f=x_0+v_0 cosθt
v_y=v_0 sinθ-a_y t
〖v_y〗^2=(v_0 sinθ)^2-a_y (y_f-y_0 )

This is not really a homework problem (it could be made to be though). I kind of made it up, inspired by a youtube math challenge problem involving parabolas, a water fountain where A = 1, R = 3, and H = 3. The solution given (h = 9/4) was based off simple math utilizing vertex form of a parabola. I wanted to find the problem just using kinematic equations. This is where I hit a roadblock (I assumed g was not necessary and treated it as an unknown). It took my a while (was dealing with 4 unknowns lol) but I eventually came up with the solution (using only kinematics). This is not normal as I'm able to do nearly all kinematic problem pretty quickly.

My question is : Would this be considered a simple problem (it wasn't for me)? If so then what technique would lead to a more elegant solution? Also, are there any solution methods involving the kinetic energy theorem?I can post my complete solution (all steps) here if necessary. Just give the word.
For now the solution is
h = H [1 - {A/(R-A)}^2]

 

[Orodruin]

I can post my complete solution (all steps) here if necessary.

Please do. These are the homework forums and you are expected to post your work even if it is not actual honework. Full solutions should generally not be given unless the OP (ie, you in this case) show that they have the full solution.

Edit: That said, I would consider it a fairly simple problem.

 

[h1a8]

I attached my solution.

 

[haruspex]

I attached my solution.

On my iPad, your equations are not visible.
But I can tell you there is a much easier way. Forget the kinetics and think about the geometry.

 

[Tom.G]

you are expected to post your work even if it is not actual honework.

In light of the below, we have a case where a typo error actually was more appropriate.

made it up, inspired by a youtube math challenge problem

 

[Orodruin]

In light of the below, we have a case where a typo error actually was more appropriate.

I blame phone + fat fingers ...

 

[h1a8]

On my iPad, your equations are not visible.
But I can tell you there is a much easier way. Forget the kinetics and think about the geometry.

The galaxy tab s7 works just fine as well as my windows laptop. I just tried my ipad and you are right. Apple needs to fix microsoft word implementation on ios. If android is capable then ios should be.
I easily performed this problem (within a few minutes) just using simple math (analytic geometry). But the challenge is just to use kinematic equations (or kinetic energy theorem equations).

 

[h1a8]

Please do. These are the homework forums and you are expected to post your work even if it is not actual honework. Full solutions should generally not be given unless the OP (ie, you in this case) show that they have the full solution.

Edit: That said, I would consider it a fairly simple problem.

I attached my solution below your post. Note: I easily performed this problem (within a few minutes) just using simple math (analytic geometry on quadratic functions). But it wasn't clear, in the beginning, how to use just only kinematic equations. Remember, I was dealing with 5 unknowns (t, v0, theta, y0, and g) and had many equations to choose from.

 

[Orodruin]

I attached my solution below your post. Note: I easily performed this problem (within a few minutes) just using simple math (analytic geometry on quadratic functions). But it wasn't clear, in the beginning, how to use just only kinematic equations. Remember, I was dealing with 5 unknowns (t, v0, theta, y0, and g) and had many equations to choose from.

Ascent:
$$
H-h = \frac{gt_1^2}2, \quad A = v_x t_1
$$
Descent:
$$
H = \frac{gt_2^2}2, \quad R-A = v_x t_2
$$
So
$$
\frac{H-h}{H} = \frac{t_1^2}{t_2^2} = \left(\frac{A}{R-A}\right)^2
$$
Solve for $h$.

 

[PeroK]

I attached my solution below your post. Note: I easily performed this problem (within a few minutes) just using simple math (analytic geometry on quadratic functions). But it wasn't clear, in the beginning, how to use just only kinematic equations. Remember, I was dealing with 5 unknowns (t, v0, theta, y0, and g) and had many equations to choose from.

Alternatively, you could set the problem up starting from the highest point. The projectile has some initial horizontal speed $v$ and the relationship between the horizontal distance and vertical height dropped is:
$$x = vt$$$$y = \frac 1 2 gt^2 = \frac 1 2 g(\frac x v)^2$$This gives the ratio of height dropped at different times:
$$\frac{y_1}{y_2} = \frac{x_1^2}{x_2^2}$$This can be adapted for your case where $x_1 = A, y_1 = H - h, x_2 = R - A, y_2 = H$.

 

[h1a8]

Ascent:
$$
H-h = \frac{gt_1^2}2, \quad A = v_x t_1
$$
Descent:
$$
H = \frac{gt_2^2}2, \quad R-A = v_x t_2
$$
So
$$
\frac{H-h}{H} = \frac{t_1^2}{t_2^2} = \left(\frac{A}{R-A}\right)^2
$$
Solve for $h$.

Alternatively, you could set the problem up starting from the highest point. The projectile has some initial horizontal speed $v$ and the relationship between the horizontal distance and vertical height dropped is:
$$x = vt$$$$y = \frac 1 2 gt^2 = \frac 1 2 g(\frac x v)^2$$This gives the ratio of height dropped at different times:
$$\frac{y_1}{y_2} = \frac{x_1^2}{x_2^2}$$This can be adapted for your case where $x_1 = A, y_1 = H - h, x_2 = R - A, y_2 = H$.

Very nice! Elegant indeed. Thank you.
Just one question :
Why does y (for ascent) doesn't have a linear term for it's initial velocity upward?
But the result still ends up as the correct solution?

Wait, Perok said set up the problem from the highest point.

That means we have to consider a descent event twice (no ascent). 1st descent event is ball falling from highest point back to released height (ball travels A distance horizontally in that event, assuming symmetry). 2nd descent event (rewinding time lol) is from highest point to landing point (ball travels R-A distance horizontally in that event).

In summary, we used 2 descending equations (to get rid of initial upward velocity term) and assumed symmetry (horizontal displacement is the same for when the ball is rising above release point as it is when the ball is falling back to the released height).

 

[Orodruin]

Alternatively, you could set the problem up starting from the highest point. The projectile has some initial horizontal speed $v$ and the relationship between the horizontal distance and vertical height dropped is:
$$x = vt$$$$y = \frac 1 2 gt^2 = \frac 1 2 g(\frac x v)^2$$This gives the ratio of height dropped at different times:
$$\frac{y_1}{y_2} = \frac{x_1^2}{x_2^2}$$This can be adapted for your case where $x_1 = A, y_1 = H - h, x_2 = R - A, y_2 = H$.

I mean, this is essentially what I did. The only reason I introduced horizontal velocity was to make it clearer that the x-displacements actually represent something directly proportional to the times.

Why does y (for ascent) doesn't have a linear term for it's initial velocity upward?

It does, but the final velocity is zero. We use this turning point as the reference point.

That means we have to consider a descent event twice (no ascent)

The descent is just the time reversal of the ascent and these equations of motion are invariant under time reversal.

 

[h1a8]

I mean, this is essentially what I did. The only reason I introduced horizontal velocity was to make it clearer that the x-displacements actually represent something directly proportional to the times.It does, but the final velocity is zero. We use this turning point as the reference point.The descent is just the time reversal of the ascent and these equations of motion are invariant under time reversal.

You guys are genius. Thanks so much. I learned something new.

 

[neilparker62]

$$v_{iy}=\sqrt{2g(H-h)}\implies t_1=\frac{\sqrt{2g(H-h)}}{g}$$
$$v_{fy}=\sqrt{2gH}\implies t_2=\frac{\sqrt{2gH}}{g}$$ Since ${v_{xi}}^2={v_{xf}}^2$:
$$\frac{g^2A^2}{2g(H-h)}=\frac{g^2(R-A)^2}{2gH}\implies\frac{A^2}{(H-h)}=\frac{(R-A)^2}{H}$$ etc.

Edit: correction as advised in post #15 - many thanks.

 

[h1a8]

Excellent Neil. But the g's in the final equation should be in the numerators (although it does not change the result).

 

[neilparker62]

Another solution using the ubiquitous vector cross product first introduced to us by PF user @kuruman in his Insights Article: How to Master Projectile Motion without using Quadratics (Equation 4) albeit not quite in the form employed in the above link.

Edit: Fixed incorrect link for "Another solution" above.

 

[kuruman]

Another solution using the ubiquitous vector cross product first introduced to us by PF user @kuruman in his Insights Article: How to Master Projectile Motion without using Quadratics (Equation 4) albeit not quite in the form employed in the above link.

Since @neilparker62 mentioned me, I feel obliged to point at yet another approach which is purely geometric in nature as @haruspex suggested. This exercise can be solved quite quickly using equations (4) and (5) derived here:
$$\begin{align}& \frac{\Delta x}{\tan\!\theta-\tan\!\varphi}=\frac{2v_{0x}^2}{g} ~~~~(4) \nonumber \\
& \tan\!\varphi=\frac{1}{2}(\tan\!\theta+\tan\!\omega)~~~~(5)\nonumber \end{align}$$The three angles are:
$\theta=$ the projection angle relative to the horizontal.
$\varphi=$ the angle of the position vector relative to the horizontal (is negative if below the horizontal.)
$\omega=$ the angle of the velocity vector relative to the horizontal (is negative if below the horizontal.)

First we note that the RHS in equation (4) is a constant of the motion. Thus we can write $$\frac{\Delta x_1}{\tan\!\theta-\tan\!\varphi_1}=\frac{\Delta x_2}{\tan\!\theta-\tan\!\varphi_2}.$$Secondly, we find the angle of projection in terms of the given quantities. To do that we use equation (5) at maximum height where $\omega=0$:$$\tan\!\varphi_1=\frac{H-h}{A}=\frac{1}{2}(\tan\!\theta+0)\implies \tan\!\theta=\frac{2(H-h)}{A}\implies\tan\!\theta-\tan\!\varphi_1=\frac{(H-h)}{A}.$$Thirdly, we find $\tan\!\varphi_2$ at the landing point, $$\tan\!\varphi_2=-\frac{h}{R}$$With $\Delta x_1=A$ and $\Delta x_2=R$, we substitute to get, $$\frac{A}{\frac{(H-h)}{A}}=\frac{R}{\frac{2(H-h)}{R}+\frac{h}{R}}.$$The rest is simple algebra which yields the answer without invoking times, velocities and the acceleration of gravity.

 

[haruspex]

Actually my geometric solution was to say that, viewed as an inverted parabola based at the highest point of the trajectory, we know the equation is $y=ax^2$.
Using the given facts,
$H-h=aA^2$
$H=a(R-A)^2$
Whence $h=H(1-(\frac A{R-A})^2)$.

 

[kuruman]

That's even quicker. I didn't think of curve-fitting.

 

[neilparker62]

Actually my geometric solution was to say that, viewed as an inverted parabola based at the highest point of the trajectory, we know the equation is $y=ax^2$.
Using the given facts,
$H-h=aA^2$
$H=a(R-A)^2$
Whence $h=H(1-(\frac A{R-A})^2)$.

I think we would have to concede that takes the cake in terms of simplicity!

 

[PeroK]

There is only one true parabola!

 

[Orodruin]

Well, it is the same solution I think... just not writing out some of the motivating steps.

 

[neilparker62]

There is only one true parabola!

Indeed !
$$y=a(x-p)^2+q$$
https://www.desmos.com/calculator/8mnuod5noh

 

[haruspex]

Indeed !
$$y=a(x-p)^2+q$$
https://www.desmos.com/calculator/8mnuod5noh

Why would you single out an algebraic form which allows for any translation and aspect ratio but not a rotation?

 

[neilparker62]

Why would you single out an algebraic form which allows for any translation and aspect ratio but not a rotation?

Essentially for simplicity and because that's one of two 'standard forms' we teach in schools here anyway.

All the same it's an interesting exercise to add rotation to the general form:

https://www.desmos.com/calculator/k1ud3ucq8x

 

[neilparker62]

With reference to the above vector diagram and derived equation, let $h_1$ be the maximum height of the projectile above launch point and $h_2$ be the maximum height of the projectile above impact point. Then:

$$u_y=\sqrt{2gh_1} ,\;\;\; v_y=-\sqrt{2gh_2}$$and $$ u_x=\frac{Ag}{\sqrt{2gh_1}}$$
Hence: $$gR=\frac{Ag}{\sqrt{2gh_1}}\left(\sqrt{2gh_1}+\sqrt{2gh_2}\right) $$
$$\implies R = A\left(1+\sqrt{\frac{h_2}{h_1}}\right)$$

The above vector diagram is sourced from this article.