Find the height of a lamp based on the velocities of projectiles

[stabby_faris]

It looks like you allowed the number 3 to intimidate you without really looking at the equations with an eye towards figuring out your approach. From
$t_1=2t_2$
$t_1=2T-t_2.$
it follows that $2t_2=2T-t_2 \implies t_2=\frac{2}{3}T.$
Can you finish it now?

yeah I know I solved it already. I took a step back and realised oh, I am already given t1 = 2t2 and then just went from there, even simpler than the fraction substitution. Thank you all for your patience

Final answer h = 2.22...m

 

[stabby_faris]

I checked back to my original approach and realised it was correct, just far more difficult to work with the algebra for it as I literally just chucked it in mathaway algebra solver to get the same h=2.22m. Thank you very much to all of you for your perspectives. This question was frankly overwhelming at times but thats a good sign, I am not stimulated enough in my classroom

 

[D H]

To summarize, you already got two equations using the notation in post #22
$t_1=2t_2$
$t_1=2T-t_2.$
Here $T$ is the time it takes object 1 to hit the floor and you an easily find it in terms of $H$ and $g$.
Add to the above an equation that says that in time $t_2$ object 2 drops to a distance $H-h$.

You now have a system of three equations and three unknowns, $t_1$, $t_2$ and $h$. Solve it.

There's no need to solve for $t_1$ and $t_2$ as the problem statement only asks for $h$.

The OP has already solved for $T$, except the OP plugged in numbers as soon as possible. Keeping it symbolic, T = \sqrt{\frac{2H}g}. Similarly, the time to fall from a height $H$ to a height $h$ with zero initial vertical velocity is t_2 = \sqrt{\frac{2(H-h)}g}. Note that this can easily be expressed in terms of $T$: t_2 = T \sqrt{1-\frac h H}.

To the OP (@stabby_faris): Note that even if one doesn't yet know values for $H$, $h$, and $g$ it is still possible to write these expressions. This is one of the many reasons to keep things symbolic.

I don't want to carry this much further as the solution is new very close.

 

[stabby_faris]

Yeah, as an exact ratio
H/h = 9/5

 

[haruspex]

Fwiw, the answers to my leading questions in post #6 are:

If the faster projectile takes time t:

• where is the other at time t?

Half way along, and since it will have fallen the same distance as the faster projectile it will be at height h.

• when will it hit the ground?

Since it will bounce elastically and finish at height h, by symmetry, that will be at time 3t/2.

• how do these times relate to h and H?

From first projectile: $H-h=\frac 12gt^2$.
From second projectile: $H=\frac 12g(3t/2)^2$.
Whence $1-\frac hH=\frac{t^2}{(3t/2)^2}=\frac 49$.

 

[stabby_faris]

Fwiw, the answers to my leading questions in post #6 are:

If the faster projectile takes time t:

• where is the other at time t?

Half way along, and since it will have fallen the same distance as the faster projectile it will be at height h.

• when will it hit the ground?

Since it will bounce elastically and finish at height h, by symmetry, that will be at time 3t/2.

• how do these times relate to h and H?

From first projectile: $H-h=\frac 12gt^2$.
From second projectile: $H=\frac 12g(3t/2)^2$.
Whence $1-\frac hH=\frac{t^2}{(3t/2)^2}=\frac 49$.

Nice compilation of the symmetry exploitation

 

[Orodruin]

For completeness: When I first looked at this I decided to work in the variable $s = \sqrt{2y/g}$ where $y$ is the distance below the original height. Both paths are linear in $s$ and just drawing straight lines in the $s$-$x$ plane you find $\sqrt{H-h} = 2\sqrt{H}/3$ graphically.

 

[PeroK]

There's no need to solve for $t_1$ and $t_2$ as the problem statement only asks for $h$.

@kuruman and I meant to solve for $h$.

 

[PeroK]

Nice compilation of the symmetry exploitation

Although there are some clever solutions presented here, this problem was solvable directly, without any insights or clever shortcuts. The bigger concern is not that you failed to spot these shortcuts, but that you failed to manage the algebra.

It's probably more helpful to you to crank through the problem in full, proving everything from first principles, and taking no shortcuts and mastering the algebra of the SUVAT equations.

 

[kuruman]

Although there are some clever solutions presented here, this problem was solvable directly, without any insights or clever shortcuts.

Final shortcut.
We have seen elsewhere that the horizontal displacement of a projectile is given by the, relatively unknown but oh so useful, equation $$\Delta x=\frac{1}{g}|\mathbf v_f\times \mathbf v_i|$$ where $\mathbf v_f$ and $\mathbf v_i$ are, respectively, the final and initial velocities. The vertical component of the velocity is a function of height and can be found quite easily. The approach is to find expressions for the horizontal displacement of each object and set them equal. Clearly, the expression for object 1 has to come in two parts because its velocity changes discontinuously upon bouncing.

Let $u=v_1$ speed of object 1 and assume that "down" is negative.

Object 2
$\mathbf v_{\text{2f}}=\{2u,-\sqrt{2g(H-h)}\}~;~~\mathbf v_{\text{2i}}=\{2u,0\}.~~$Then,$$
\Delta x=\frac{1}{g}|\mathbf v_{\text{2f}}\times \mathbf v_{\text{2i}}|=
\frac{1}{g} \left | \{2u,-\sqrt{2g(H-h)}\}\times \{2u,0\} \right |=\frac{2u}{g}\sqrt{2g(H-h)}.$$Object 1
For the "going down" leg (a),
$\mathbf v_{\text{1f}}^{(a)}=\{u,-\sqrt{2gH}\}~;~~\mathbf v_{\text{1i}}^{(a)}=\{u,0\}.~~$ $$
\Delta x_a=\frac{1}{g} \left | \{u,-\sqrt{2gH}\}\times \{u,0\} \right |=\frac{u}{g}\sqrt{2gH}.$$ For the "coming back up" leg (b),
$\mathbf v_{\text{1f}}^{(b)}=\{u,\sqrt{2g(H-h)}\}~;~~\mathbf v_{\text{1i}}^{(b)}=\{u,\sqrt{2gH}\}.$
$$
\Delta x_b=\frac{1}{g} \left | \{u,\sqrt{2g(H-h)}\}\times \{u,\sqrt{2gH}\} \right |=
\frac{u}{g}\left[\sqrt{2gH}-\sqrt{2g(H-h)} \right].$$ Putting the expressions together, $$\begin{align} &
\Delta x=\Delta x_a+\Delta x_b \nonumber \\
& \frac{2\cancel u}{\cancel g}\sqrt{\cancel{2g}(H-h)}=\frac{\cancel u}{\cancel g}\sqrt{\cancel{2g}H}+\frac{\cancel u}{\cancel g}\left[\sqrt{\cancel{2g}H}-\sqrt{\cancel{2g}(H-h)} \right] \nonumber \\
& 3\sqrt{H-h}=2\sqrt{H} \nonumber \\ & H-h=\frac{4}{9}H\implies h=\frac{5}{9} H.
\nonumber \end{align}$$No quadratics, no use of symmetry, no kinematic equations, no time-of-flight considerations.

 

[kuruman]

We would be remiss not to add a geometric proof based on velocity vs. time plots. The figure on the right shows the vertical velocity of the two objects as a function of time.

The plot for object 2 is a straight (red) line from O to D in time $~t_2=\text{(OA)}.$ The plot for object 1 overlaps that of object 2 and is extended to point E. At time $t_E=\text{(OB)}$ the vertical velocity reverses direction discontinuously to point H and then drops to point K at time $t_1=\text{(OC)}.$

Since $t_1=2t_2=\text{(OA)}+\text{(AC)}$, it follows that $\text{(OA)}=\text{(AC)}.$ Now right triangles HJK an EFD have their sides parallel and this makes them similar. The equality $\text{(HJ)} =\text{(EF)}$ makes them congruent. Thus, $\text{(JK)} =\text{(DF)}\implies \text{(AB)}=\text{(BC)}.$

From this last result it follows that $\text{(AB)}=\frac{1}{2}\text{(OA)}$ and that $\text{(OB)}=\text{(OA)}+\text{(AB)}=3\text{(AB)}.$

From similar triangles EFD and EBO, $$ \begin{align} & \frac{\text{(DF)}}{\text{(OB)}}=\frac{\sqrt{2gH}-\sqrt{2g(H-h)}}{\sqrt{2gH}}.\nonumber \\
& \frac{\cancel{\text{(AB)}}}{3\cancel{\text{(AB)}}}=\frac{\sqrt{H}-\sqrt{H-h}}{\sqrt{H}} \nonumber \\
& 3\sqrt{H-h}=2\sqrt{H}\implies h=\frac{5}{9}H. \nonumber \end{align}$$