Two balls, dropped with a delay of ##\Delta t##, meet after rebound

[brotherbobby]

Homework Statement

Two balls are dropped from a top of a cliff at a time interval of $\Delta t = 2\;\text{s}$. The first ball hits the ground and rebounds elastically, essentially reversing its direction instantly without losing speed. It collides with the second ball at a height of $\text{55 m}$ above the ground. How high is the top of the cliff?

Relevant Equations

For uniformly accelerated motion under gravity, taking $g = +9.8\;\text{m/s}^2$, we have the position after a time $t$, $y(t) = y_0+v_0t-\frac{1}{2}gt^2\;\text{(I)}$ and the velocity $v(t)=v_0-gt\; \text{(II)}$. The velocity can also be expressed as a funtion of the position $y$ from the origin, $v^2(y)=v^2_0-2gy\;\text{(III)}$.

I draw a series of images of the problem situation. In (a), the first ball in green is (just) dropped at time $t=0$. In (b), the second ball in red is dropped at time $t=2\;\text{s}$. In (c), the first ball rebounds after a total time of falling down $t_d$. At this position, its velocity $v_1(t_d)$ is the same in either direction. In (d), the balls meet after a total time of flight $t$ at a height $h$. The sign convention is positive $+$ in the upwards direction and all distances measured from the ground as the origin $\color{blue}{\textbf{O}}$.

For the second ball at collision point, $y_2(t) = h = H - \frac{1}{2}g(t-2)^2$, from which we have its equation of motion,
\begin{equation*}
\color{red}{H-h =\frac{1}{2}g(t-2)^2} \quad\quad (2)
\end{equation*}
For the first ball at the time colliding, we can again write $y_1(t) = h$, but we have to be careful here because the motions, first down and then up, take place after a change of acceleration to infinity at the point where the ball hits the ground after a time $t_d$. We have to do each part separately.
Let $t_u$ be the time the first ball takes from the time it rebounds to collide with the second. So we have at the time of collision, $t = t_d+t_u$. We can find the time for the downward motion easily, $t_d=\sqrt{\frac{2H}{g}}$. Hence we have $t_u = t-\sqrt{\frac{2H}{g}}$. The velocity following this downward motion $v_1(t_d) = \sqrt{2gH}$. For the subsequent upward motion of the ball, $h = v_1(t_d)t_u-\frac{1}{2}gt_u^2$. Hence the equation of motion of the first ball on its way up,
\begin{equation*}
\color{ForestGreen}{h = \sqrt{2gH}\left(t-\sqrt{\frac{2H}{g}}\right)-\frac{1}{2}g\left(t-\sqrt{\frac{2H}{g}}\right)^2} \quad\quad (1)
\end{equation*}
From equation $(2)$, we have $t=\sqrt{\frac{2(H-h)}{g}}+2$.
Inserting this value of the time $t$ in equation $\text{1}$, we get
$$\small{h = \sqrt{2gH}\left( \sqrt{\frac{2(H-h)}{g}} +2-\sqrt{\frac{2H}{g}}\right)-\frac{1}{2}g\left\{ \frac{2(H-h)}{g} +2 -\sqrt{\frac{2H}{g}}\right\}^2}$$
Upon expanding the equation above, it looks hopeless trying to express $H$ as a function of $h$.

Request : A hint to solve the problem.

 

[DrClaude]

Request : A hint to solve the problem.

If all you want is to solve the problem, introduce numbers earlier. I would start by considering only the motion of the first ball to calculate the time of the collision.

 

[brotherbobby]

If all you want is to solve the problem, introduce numbers earlier. I would start by considering only the motion of the first ball to calculate the time of the collision.

$h = \text{55 m}$, given. That's the only thing I have left as a variable. Let me see if putting 55 m helps.

 

[PeroK]

$h = \text{55 m}$, given. That's the only thing I have left as a variable. Let me see if putting 55 m helps.

That's a very high bounce! It can't be a beach at the bottom of the cliff.

 

[PeroK]

Request : A hint to solve the problem.

The motion of the first ball from $h$ to the ground and bouncing back up to $h$ is symmetrical. I.e. it takes as long to fall as to bounce back up. Each leg of the motion must take $\frac{\Delta t}{2} = 1s$.

 

[Delta2]

The motion of the first ball from $h$ to the ground and bouncing back up to $h$ is symmetrical. I.e. it takes as long to fall as to bounce back up. Each leg of the motion must take $\frac{\Delta t}{2} = 1s$.

This seems a nice and neat conclusion but why do you set Δt=2=the time interval between the two balls launch events?

 

[PeroK]

This seems a nice and neat conclusion but why do you set Δt=2=the time interval between the two balls launch events?

Because the second ball is where the first ball was two seconds ago. The second ball will have $1s$ more before hitting the ground and a further $1s$ to bounce back up to where the balls passed each other.

For example, if $h = 5m$, then (with $g = 10m/s^2$), the first ball will take $1s$ to hit the ground and $1s$ to bounce back up - arriving back at the top of the cliff just as the second ball is about to be dropped. And, in this case the height of the cliff must be $5m$.

 

[Delta2]

Because the second ball is where the first ball was two seconds ago

Yes right, let me explain it a bit better e hehe.

Since the second ball is where the first ball was two second ago , at the place they meet 55m above ground, means that the first ball was there not only now but also two seconds ago, and since the up and down time are equal, it took the first ball 1 second to hit the ground and 1 second to rise up to 55m again (where it meet the second ball).

This is a rather heavy intuitional explanation and it might not satisfy everyone's intuition. If we gonna do it formally with math and equations how are we gonna do it ?

 

[PeroK]

If we gonna do it formally with math and equations how are we gonna do it ?

The symmetry is inherent in the equations of motion; and by conservation of mechanical energy.

 

[Delta2]

The symmetry is inherent in the equations of motion; and by conservation of mechanical energy.

I throw you the ball and you throw it back to me lol I mean I was trying to make you do the formal math for this conclusion e hehe. It doesn't seem so easy to me, and I don't know if we have to "throw the ball to the OP" I mean say to the OP to prove this formally, or there is another nice neat and more formal way of solving this without relying on this intuitive conclusion.

 

[PeroK]

I throw you the ball and you throw it back to me lol I mean I was trying to make you do the formal math for this conclusion e hehe. It doesn't seem so easy to me, and I don't know if we have to "throw the ball to the OP" I mean say to the OP to prove this formally, or there is another nice neat and more formal way of solving this without relying on this intuitive conclusion.

There's no intuition in my solution. Looking for symmetry in a solution is fundamental.

You seem to be mandating an approach that ignores the physics of the problem and just blindly manipulates equations?

 

[Delta2]

There's no intuition in my solution. Looking for symmetry in a solution is quite fundamental.

If you ask me all the symmetry explanations whether it is classical electromagnetism or classical mechanics/kinematics seem very intuitive to me and some times very vague too

You seem to be mandating an approach that ignores the physics of the problem and just blindly manipulates equations?

YES! Of course we cant ignore completely the physics , the starting equations would come from the physics of the problem.

 

[Delta2]

Ok ok maybe not exactly intuitive, but qualitative-intuitive. Something that Faraday used to do in the 1830-40s-50s when he was laying the foundations of classical EM .

 

[erobz]

I think I found an algebraic path, but its messy.

I start with the finding the time for the first ball to bounce, and then the position, velocity of the second ball at that time. After that I rewrite the EOM for each ball in terms of $g,H,$, and $\Delta t$( the delay) and restart the clock, to find the time they will meet $t_x$. The position of the ball that has bounced is $h$ at that time, and the position of the second ball is $H-h$.

You can then solve for $t_x$ in terms of $g,H,h,\Delta t$

Then plug $t_x$ back into the position equation of the ball that has bounced.

And you are left with an equation in $g,H,h,\Delta t$ with the only unknown $H$. I didn't carry on to see if its analytical (looks quite messy - but I have been shocked many times on looks before too )

EDIT: I didn't realize the OP already found an expression in these variables. Whoops!

 

[erobz]

for your solution in the OP I think make the substitution:

$U= \frac{2(H-h)}{g} + \Delta t - \sqrt{ \frac{2H}{g}}$

And solve the resulting quadratic in $U$

Then you will be left with something like:

$$ \left( \frac{2(H-h)}{g} + \kappa \right) = \pm \frac{1}{g} \sqrt{ \beta + 2 g h } $$

Where $\kappa$ and $\beta$ are just constants.

Then square both sides and I think you will be left with a quadratic to solve in $h$.Never mind, wrong variable...

 

[PeroK]

I think I found an algebraic path, but its messy.

The alternative is to prove (using the SUVAT equations), what must be true by symmetry. An object falling from an initial height $H$, reaches a height $h$ at time $t_h$, where:
$$h = H - \frac 1 2 gt_h^2$$$$\Rightarrow \ t_h = \sqrt{\frac{2(H-h)}{g}}$$The time to hit the ground, $t_0$, is given by:$$t_0 = \sqrt{\frac{2H}{g}}$$So, the object is at height $h$ a time $\Delta t$ before it hits the ground, where:$$\Delta t = t_0 - t_h = \sqrt{\frac{2H}{g}} - \sqrt{\frac{2(H-h)}{g}}$$The speed of the object when it hits the ground becomes the initial velocity for the rebound. Using $v^2 - u^2 = 2as$, we have:
$$v_0 = \sqrt{2gH}$$The ball rebounds to a height $h$ at time $T_h$ after the rebound where:
$$h = v_0T_h - \frac 1 2 g T_h^2 $$Solving the quadratic for the first instance of $T_h$ gives:
$$T_h = \sqrt{\frac{2H}{g}} - \sqrt{\frac{2(H-h)}{g}} = \Delta t$$And the result has been cranked out for those without the insight to see the physical symmetry inherent in the scenario. And, with a clear conscience, you can now solve the problem zee kwick vay.

 

[Delta2]

Oh come on @PeroK what you did is to prove that the uptime equals the downtime, I want a formal proof that the total uptime+downtime equals the time interval between the dropping of the two balls

 

[erobz]

And, with a clear conscience, you can now solve the problem zee kwick vay.

I'm a glutton for punishment though...

Last try:

let $U = t - \sqrt{\frac{2H}{g}}$ from eq(1)

solve (1) for $U$

$$ u^2 - \frac{2}{g} \sqrt{2gH} u + \frac{2}{g}h = 0 $$

$$ \implies u = \frac{2}{g} \sqrt{2gH}+ \sqrt{ \frac{8}{g} \left( H - h \right)} $$

I checked that the negative root was not a valid solution because it must be that $u > 0$.

so we have:

$$ t - \sqrt{ \frac{2H}{g}} = \frac{2}{g} \sqrt{2gH}+ \sqrt{ \frac{8}{g} \left( H - h \right)} $$

sub for $t$ you found from 2

$$ \sqrt{ \frac{2}{g} \left( H - h \right)} + \Delta t - \sqrt{ \frac{2H}{g}} = \frac{2}{g} \sqrt{2gH}+ \sqrt{ \frac{8}{g} \left( H - h \right)} $$

Then massage what is under the roots

$$ \frac{1}{2}\sqrt{ \frac{8}{g} \left( H - h \right)} + \Delta t - \sqrt{ \frac{2H}{g}} = \frac{2}{g} \sqrt{2gH}+ \sqrt{ \frac{8}{g} \left( H - h \right)} $$

$$-\frac{1}{2}\sqrt{ \frac{8}{g} \left( H - h \right)} = \frac{2}{g} \sqrt{2gH}+\sqrt{ \frac{2H}{g}} - \Delta t $$

more massaging of the RHS:

$$-\frac{1}{2}\sqrt{ \frac{8}{g} \left( H - h \right)} = \frac{2}{g} \sqrt{2gH}+\frac{1}{g}\sqrt{ 2gH} - \Delta t $$

$$ -\frac{1}{2}\sqrt{ \frac{8}{g} \left( H - h \right)} = \frac{3}{g} \sqrt{2gH} - \Delta t $$

Square both sides:

$$ \frac{2}{g}\left(H - h \right) = \frac{18}{g}H - \frac{6 \Delta t}{g} \sqrt{2gH} + {\Delta t}^2 $$

Isolate the root on the RHS again, square both sides and solve the resulting quadratic in $H$.

At least I hope I haven't bungled again.

 

[Delta2]

ok well, what zee kwick vay means , i google it but dont seem to find anything, at least not in the first few pages.

 

[PeroK]

@brotherbobby if you use my hint (expanded in post #16), then it's not too hard to get an expression for $H$ in terms of $h, g$ and $\Delta t$. Alternatively, you can break the problem down into two steps by calculating the velocity at impact (based on post #16). And, from that calculate $H$.

 

[brotherbobby]

That's a very high bounce! It can't be a beach at the bottom of the cliff.

Imagine the experiment taking place in the grand canyon. Rocks all round. Not sure though whether you call them hills, mountains or cliffs over there.

 

[berkeman]

ok well, what zee kwick vay means , i google it but dont seem to find anything, at least not in the first few pages.

LOL, it looks like it's a language translation thing. "Zee kwick vay" is a humorous adaption of "the quick way", spoken in English with a heavy German (or similar) accent.

 

[PeroK]

Imagine the experiment taking place in the grand canyon. Rocks all round. Not sure though whether you call them hills, mountains or cliffs over there.

It would end up in the Colorado river!

 

[PeroK]

LOL, it looks like it's a language translation thing. "Zee kwick vay" is a humorous adaption of "the quick way", spoken in English with a heavy German (or similar) accent.

As spoken by Dr Einstein, played by Peter Lorre in Arsenic and Old Lace. Griffiths quotes it in his QM book: ze slow vay is zo messy!

 

[kuruman]

One can easily set up the relevant equation by using a speed vs. time graph. Shown below is the speed of the first ball in red and the second ball in blue. The point of intersection C is the collision point at the time when the speeds are equal as required by energy conservation. The velocities are in opposite directions, of course.

The height of the cliff $H$ is the area of triangle OAB. The base of this triangle is the time it takes the first ball to hit the ground $OB=t_f=\sqrt{\frac{2H}{g}}.$ Triangle EAD is similar to OAB. Therefore, $$\frac{(AD)}{(AB)}=\frac{(ED)}{(OB)}\implies \frac{\sqrt{2gH}-\sqrt{2g(H-h)}}{\sqrt{2gH}}=\frac{\frac{1}{2}t_0}{\sqrt{\frac{2H}{g}}}.$$

On edit
Edited to fix the incorrectly applied "area under the curve" equation.

 

[brotherbobby]

The motion of the first ball from $h$ to the ground and bouncing back up to $h$ is symmetrical. I.e. it takes as long to fall as to bounce back up. Each leg of the motion must take $\frac{\Delta t}{2} = 1s$.

I am the OP of the problem; sorry for coming in late. I was just as surprised by @PeroK's hint as the others, not least because I had no clue. Despite its beauty, for beginners like me to this kind of thinking, it requires some explaining, if only to convince oneself.

Let me start with the statement of the problem as a refresher.

Hint : $\text{The first ball takes exactly 1 s from the time it rebounds to where it meets the second ball.}$

The first ball is always a time of 2 s ahead of the second ball. However, this does not mean, as a digression, that their distance of separation remains the same. But whatever mark along the cliff the first ball passes, the second ball also will, 2 s later. The first ball passed the position of collision (which happened later on) which is given to be at a height of $h = 55\;\text{m}$ above the ground. Since it rebounded with the same speed with which it struck the ground, it would climb back to the height $h$ taking the same time that it took to descend. Again, as a digression, note that this is true for all subsequent heights climbed by the first ball, if it was allowed to, due to the symmetry of free fall motion. However, something special happens at the given height $h\;-\;$ when the first ball reaches that height, the second ball meets it to collide. Since the first ball was 2 s ahead of the second ball, it must have taken the first ball a time of 1 s to travel to and from the height $h$ to the ground!

Attempt : With the hint above, the solution of the problem is straightforward. The time taken by the first ball (in green) to meet the second (in red) after collision is what I call $t_u = \frac{\Delta t}{2} = \frac{2}{2} = 1\;\text{s}$. Using $y(t) = y_0+v_0t-\frac{1}{2}gt^2$ for this rebound motion, we have $\small{h = v_1(t_d)t_u-\frac{1}{2}gt_u^2\Rightarrow 55 = v_1(t_d)-5}$, taking $g = 10\;\text{m/s}^2$. This leads to $v_1(t_d) = 60\,\text{m/s}$, as the speed with which the first ball hits the ground and rebounds after a time of flight $t_d$ downward. We need not find this time, because using the third $\text{(III)}$ of the relevant equations given above, viz. $v^2(y)=v^2_0-2gy\;\text{(III)}$, $v_1^2(t_d) =v_0^2-2gH\Rightarrow H = \frac{v_1^2(t_d)}{2g}=\frac{60^2}{2\times 10} = \boxed{180\; \text{m}}\color{green}{\large{\checkmark}}$, as $v_0=0$ due to fall from rest.

While this matches the answer in the text, there are at least two questions that remain to me, as doubts.

1. Can this method of solution be employed if the first ball rebounded with a fraction of the velocity with which it struck the ground? I am inclined to say "no". The symmetry of the problem is lost. And if that's the case, the mathematical solution that I attempted earlier on and duly failed at, would be the only way out. However, it would be even harder, obviously, to do so.

2. Is motion in a straight line symmetric between rise and fall if acceleration is not uniform? I am inclined to say "yes", if acceleration has a spatial dependence, only; i.e. $a = a(x)$. But "no", if there was a temporal dependence too.

I welcome being corrected on the two points above. It is time to do the problem again in the brute way now, using the equations of motion and ignoring the symmetry of the situation.

 

[PeroK]

1. Can this method of solution be employed if the first ball rebounded with a fraction of the velocity with which it struck the ground? I am inclined to say "no". The symmetry of the problem is lost. And if that's the case, the mathematical solution that I attempted earlier on and duly failed at, would be the only way out. However, it would be even harder, obviously, to do so.

That would be a more challenging problem. The same overall methods could be used, but the complicated algebra would be hard to avoid.

2. Is motion in a straight line symmetric between rise and fall if acceleration is not uniform? I am inclined to say "yes", if acceleration has a spatial dependence, only; i.e. $a = a(x)$. But "no", if there was a temporal dependence too.

I welcome being corrected on the two points above. It is time to do the problem again in the brute way now, using the equations of motion and ignoring the symmetry of the situation.

If the acceleration decreased with height, there would still be symmetry (using the energy conservation argument). The symmetry insight is even more valuable. In fact, in general, as the problems become more advanced these insights become critical to keep the complexity under control.

 

[brotherbobby]

let U=t−2Hg from eq(1)

sub for t you found from 2

Sorry to bother @erobz . I am stuck early on in the solution you provided in post #18, unsure what you meant by equations (1) and (2). Can you clarify?

 

[erobz]

Sorry to bother @erobz . I am stuck early on in the solution you provided in post #18, unsure what you meant by equations (1) and (2). Can you clarify?

Fair warning I haven't taken it the full way through, because the last step is still quite laborious\tedious.

The equations I'm referring to are the equations you found in the OP. You have them labeled $(1)$ in green, and $(2)$ in red.

 

[brotherbobby]

Hi @erobz, can you check your working that I put below which you did in post #18 above? I think you are off by a factor of 2. I put your workings on the right, with the mistakes in red ink.

Let me copy and paste what I am getting. I mark the final expression for $u$ in orange.

What's more, I did not understand your reason for keeping the positive ($+$) sign for $u$. I copy and paste your reason to the right.

Indeed, $u = t-\sqrt{\frac{2H}{g}}$ and since $t>\sqrt{\frac{2H}{g}}\Rightarrow u>0$. But is that the reason to keep the $+$ sign?
See my reasoning below to the right on dark page.

What I am saying basically is that the second term is less than the first in the expression for $u$. So keeping the $-$ sign continues to keep $u>0$.

Many thanks.

 

[erobz]

Hi @erobz, can you check your working that I put below which you did in post #18 above? I think you are off by a factor of 2. I put your workings on the right, with the mistakes in red ink.
View attachment 339076

Let me copy and paste what I am getting. I mark the final expression for $u$ in orange.

View attachment 339077

View attachment 339078What's more, I did not understand your reason for keeping the positive ($+$) sign for $u$. I copy and paste your reason to the right.

Indeed, $u = t-\sqrt{\frac{2H}{g}}$ and since $t>\sqrt{\frac{2H}{g}}\Rightarrow u>0$. But is that the reason to keep the $+$ sign?
See my reasoning below to the right on dark page.

View attachment 339081
What I am saying basically is that the second term is less than the first in the expression for $u$. So keeping the $-$ sign continues to keep $u>0$.

Many thanks.

Yes, you are correct on both accounts. I goofed on the algebra, and one of the solutions cannot be ruled out by negative time.

So if both solutions are positive for $u$, must we then take the smallest of the two as the desired solution? ( i.e. $u = \frac{1}{g}\sqrt{2gH} - \sqrt{\frac{2(H-h)}{g}}$ ) or do we just work with both solutions $\pm$ ( I'm leaning towards the former)?

 

[brotherbobby]

@erobz, you were right. I think I have solved it, though a question lingers. Since it has been sometime that I have posed the problem, let me state things from the beginning to refresh.

Attempt : Taking $g = 9.8\:\text{m/s}^2$, we have the equation of motion of the first ball as (see my post #1) shown on the right. Here $h$ is the height where the balls meet, $t$ is the time when they meet (from the start, when the first ball was dropped). The equation of motion of the second ball (see my post #1) is also shown on the right. The question is now to find the height of the cliff H (=?).

In $(1)$, let us put $\color{blue}{u = t-\sqrt{\frac{2H}{g}}}\qquad \color{brown}{(3)}$
This, after some algebra, we obtain $\small{u^2-\sqrt{\frac{8H}{g}}u+\frac{2h}{g}=0\Rightarrow u = \sqrt{\frac{2H}{g}}\pm\sqrt{\frac{2(H-h)}{g}}=\sqrt{\frac{2}{g}}(\sqrt H\pm \sqrt{H-h})}$
Substituting from $(3)$ for $u=t-\sqrt{\frac{2H}{g}}=\sqrt{\frac{2}{g}}(\sqrt H\pm \sqrt{H-h})\Rightarrow t = \sqrt{\frac{8H}{g}}\pm \sqrt{\frac{2(H-h)}{g}}$.
From equation $(2)$, we have $\sqrt{\frac{2(H-h)}{g}}=t-2\Rightarrow \color{blue}{t = \sqrt{\frac{8H}{g}}\pm (t-2)}\qquad \color{brown}{(4)}$

There are two possibilities to $(4)$ as evidenced by the $\pm$ sign.

$1. \;\mathbf{(-)\,\text{sign}}$ :

This yields from $(4)$, $2t-2 = \sqrt{\frac{8H}{g}}\Rightarrow t= \sqrt{\frac{2H}{g}}+1\Rightarrow t-\sqrt{\frac{2H}{g}}=1\quad(=u)$. Putting this value of $t$ in $(1)$, we have,
$\small{h = \sqrt{2gH}-\frac{1}{2}g\Rightarrow h+\frac{1}{2}g = \sqrt{2gH}\Rightarrow 55+5=\sqrt{20H}\quad\text{(putting values here)}\Rightarrow 60^2=20H\Rightarrow} \boxed{H = 180\,\text{m}}\quad\large{\color{green}{\checkmark}}$

This agrees with the answer.

$2. \;\mathbf{(+)\,\text{sign}}$ :

This yields from $(4)$, $t=\sqrt{\frac{8H}{g}}+t-2\Rightarrow \sqrt{\frac{8H}{g}} = 2\Rightarrow H = \frac{4g}{8}= 5\,\text{m}\quad\huge{\color{red}{\times}}$.

Of course this answer is incorrect, as the problem states that the ball collide at $h = 55\,\text{m}$. But it should mean something that I am not able to figure out.

 

[erobz]

Good job. Thanks for finishing it out zee messy vay!

But it should mean something that I am not able to figure out.

As for the extraneous solution, I'm not concerned because temporally it makes sense to exclude the positive root solution of $u$, because the solution corresponding to the negative root $u$ must be the first collision temporally (if there were indeed two).

If it were the other way around and we found the solution corresponding to the positive root (coming after the other solution in time), I would be concerned...

 

[brotherbobby]

Glad the thread remains open. I had a question for @PeroK, if you're here.

In post# 9, you said that

The symmetry (of free fall - my addition) is inherent in the equations of motion

I am trying to investigate the matter but I am stuck somewhere I want to tell you about. Let me begin simply.

$(1)$. If an object is thrown vertically up, the time of rise = time of fall as well we the initial velocity of projection = final velocity. [I can prove both of these - they are elementary].

$(2)$. From (1) above, we can show that the time to climb to a height $y$ = time to fall from the height $y$ to the ground, even though at the height $y$ the ball has a finite speed. [Contrast with (1) where the speed at the highest point is 0].

$(3)$. Is it also true that for a ball thrown vertically up and which rises to a maximum height $h$, the time to scale a height $y$ is the same as the time that the ball dropped from the height $h$ will take to fall a distance $y$ down?

[ I cannot show $(3)$ either using symmetry arguments or algebra using the kinematic equations ].

Many thanks.

 

[PeroK]

$(3)$. Is it also true that for a ball thrown vertically up and which rises to a maximum height $h$, the time to scale a height $y$ is the same as the time that the ball dropped from the height $h$ will take to fall a distance $y$ down?

[ I cannot show $(3)$ either using symmetry arguments or algebra using the kinematic equations ].

Many thanks.

That can't be the same because the initials speeds are different.

For an elastic ball, the time to fall from height $h$ is the same as the time to bounce back up to height $h$.

First, we have the same acceleration due to gravity, $g$. Second, we have the same distance. Third, we have the same final speed in the first case as the initial speed in the second case.

This implies that the initial speed in the first case (speed at height $h$) is the same as the final speed in the second case (also height $h$). This can also be seen directly from conservation of mechanical energy. Note that conservation of energy also covers the case of a varying gravitational force/potential.

For constant acceleration, you can use the kinematic equations to show that the time is the same up and down.