Catching a loaf of bread thrown vertically at a window

[brotherbobby]

Homework Statement

Jerry throws a loaf of bread to George, who is in front of an open window whose lower edge is at height $H$ and whose upper edge is at $H + h$.

(i) What should be the initial upward speed of the bread if it is to just make it to the bottom of the window?

(ii) If it is to just make it to the top of the window?

(iii) How long does George have to grab the loaf in the second case?

Relevant Equations

The well known equations of kinematics are all I suspect are required, adjusted a bit for a gravitational field where the uniform acceleration $a_0=-g=-9.8\rm{m/s^2}$. Here $x_(t_0)=x_0$ and $v(t_0)=v_0$ and the starting time $t_0$ could be put $0$ for convenience.
\begin{align}
&v(t)=v_0+a_0(t-t_0)\\
&x=x_0+v_0(t-t_0)+\dfrac{1}{2}a_0(t-t_0)^2\\
&v^2(x)=v^2_0+2a_0(x-x_0)\\
\end{align}

I start by copying and pasting the problem statement as it appears in the text.

Let's have a diagram for the problem. A loaf of bread, drawn in brown, is thrown up from the ground with two different speeds $(v_0)_{1}$ and $(v_0)_{2}$ such that it just reaches the bottom and top of the window, drawn in red. The window has a height $h$ and its bottom is at a height $H$ from the ground, taken to be the origin O.
(i) If the loaf is to just reach the botton of the window labelled as $1$, its velocity there $v(1)=0$. Using equation $(3)$ from above, $v^2(1)=(v^2_0)_1-2gH\Rightarrow\boxed{(v_0)_1=\sqrt{2gH}}\quad\color{green}{\large{\checkmark}}\qquad(4)$

(ii) Likewise, for the loaf to just reach the top of the window at a height $(H+h)$ from the ground, we find that the initial speed should be $\boxed{(v_0)_2=\sqrt{2g(H+h)}}\quad\color{green}{\large{\checkmark}}\qquad(5)$

(iii) How much time will the loaf in (ii) above take to cross the height of the window $h$? (This is where my answer doesn't match that in the text)

If $t_1$ be the time to reach the bottom of the window and $t_2$ the time to reach the top, the required time $\Delta t = t_2-t_1$. Using equation $(2)$ from the relevant equations above for the bottom, we get $H=(v_0)_2t_1-\frac{1}{2}gt_1^2\Rightarrow t_1^2-\frac{2(v_0)_2}{g}t_1+\frac{2H}{g}=0\Rightarrow t_1=\frac{\frac{2(v_0)_2}{g}\pm\sqrt{\frac{4(v_0^2)_2}{g^2}-\frac{8H}{g}}}{2}\Rightarrow t_1=\dfrac{(v_0)_2}{g}-\sqrt{\dfrac{(v_0^2)_2}{g^2}-\dfrac{2H}{g}}$
For the time to reach the top of the window with initial speed $(v_0)_2$, we use equation $(1)$ above, noting that at the top of the window, $v(2)=0$ to obtain the time $t_2= \dfrac{(v_0)_2}{g}$.
Thus the time the loaf takes to cross the height $h$ of the window :
$\Delta t=t_2-t_1 = \sqrt{\frac{(v_0^2)_2}{g^2}-\frac{2H}{g}}$. Substituting from the value found in $(5)$ above, $\Delta t= \sqrt{\frac{2(H+h)}{g}-\frac{2H}{g}}\Rightarrow \boxed{\Delta t=\sqrt{\dfrac{2h}{g}}}\quad\color{red}{\huge{\times}}$.

Let me copy and paste the answer from the text.


Request : Where did I go wrong in my calculation in part (iii)?

 

[TSny]

If $t_1$ be the time to reach the bottom of the window and $t_2$ the time to reach the top, the required time $\Delta t = t_2-t_1$.

George can catch the bread on its way back down as well as on the way up.

 

[brotherbobby]

Yes. So that should mean that the time available to George $\Delta t = 2\sqrt{\dfrac{2h}{g}}=\boxed{\sqrt{\dfrac{8h}{g}}}$.
Thanks.

 

[TSny]

There is a simpler way to get the time interval for the bread to pass the window. It takes the same time to pass on the way up as on the way down. It is easy to calculate the time for the bread to fall from the top of the window to the bottom of the window.

 

[brotherbobby]

There is a simpler way to get the time interval for the bread to pass the window. It takes the same time to pass on the way up as on the way down. It is easy to calculate the time for the bread to fall from the top of the window to the bottom of the window.

Yes, the time to fall from the top of the window to the bottom is just $\sqrt{\dfrac{2h}{g}}$. This should be the same going up leading to the final answer.

 

[Hill]

In the answer from the text,

what is the $s$?

 

[brotherbobby]

In the answer from the text,
View attachment 339846
what is the $s$?

Seconds, I suppose.

 

[jbriggs444]

Seconds, I suppose.

Making the book answer technically incorrect.

The inputs were not given with units. The calculation was not done with units. The computed result should be a quantity of time, not a number with units.

 

[Hill]

Seconds, I suppose.

If it is so, then it is wrong, isn't it?

 

[brotherbobby]

Making the book answer technically incorrect.

The inputs were not given with units. The calculation was not done with units. The computed result is a quantity of time, not a number with units.

I agree with you.

 

[brotherbobby]

If it is so, then it is wrong, isn't it?

There was simply no reason to give $\text{s}$ as units. No units were given anywhere in the question.
By the way, this is R. Shankars (2019) Fundamentals of Physics (I).

 

[PeroK]

What other unit of time would be used in this case? We're not likely to measure the time in years!

 

[Hill]

Somewhat OT, this question has reminded me of a question I've received some time ago from a friend. I copy and paste the original and an English translation of it:

Bob got an electronic clock which shows time with a precision of a hundredth of a second. As he was moving down on an escalator, Bob threw the clock up and noticed that at the top of its trajectory the clock showed 11:32:45:81. His teacher Mary was moving up on the escalator at the same time, and she noticed that the clock showed 11:32:45:74 at the top of its trajectory. Find the speed of the escalators, given that they move with the same speed, at the angle of $30^o$ to horizon. Ignore the air friction. Take $g = 10 m/s^2$.

Spoiler: My answer

0.7 m/s

 

[jbriggs444]

What other unit of time would be used in this case? We're not likely to measure the time in years!

Microfortnights

 

[brotherbobby]

A very good problem, so thanks to @Hill . My problem was trivial in comparison.
As the OP of this thread, I should try and do it.

Bob and Mary see a different time for the clock peaking its trajectory because $\text{trajectory top}\rightarrow v_{\text{max}}=0$ happens at a different time for each of them, given their own relative motions. Problem is, the same difference would remain with the speed that Bob threw the clock, say some $v_0$. Won't the two differences compensate one another and result in the same time for maximum height, in contradiction to what the problem states? I think I should try it out.

Is the maximum height attained by the clock invariant for both? It should be, given that it's a length.

 

[Hill]

Is the maximum height attained by the clock invariant for both? It should be, given that it's a length.

I don't think so.

 

[PeroK]

Somewhat OT, this question has reminded me of a question I've received some time ago from a friend. I copy and paste the original and an English translation of it:
View attachment 339848
Bob got an electronic clock which shows time with a precision of a hundredth of a second. As he was moving down on an escalator, Bob threw the clock up and noticed that at the top of its trajectory the clock showed 11:32:45:81. His teacher Mary was moving up on the escalator at the same time, and she noticed that the clock showed 11:32:45:74 at the top of its trajectory. Find the speed of the escalators, given that they move with the same speed, at the angle of $30^o$ to horizon.

If Mary was Bob's cousin, this would be a good example of relative motion!

 

[brotherbobby]

Let me post a simpler problem than the one in post #13 above.

Problem : A ball is thrown with a velocity $v_0$ vertically up from the ground. An elevator was moving up from the ground with a uniform velocity $v_E \;(<v_0)$ at the same instant. What is the maximum height $H'$ that the ball attains in the elevator's frame and when?

Solution : The ball has an initial velocity of $v'_0=v_0-v_E$ in the elevator's frame. Using the velocity-displacement equation ($(3)$ above), the maximum height attained by the ball in the elevator's frame is $\boxed{H'=\dfrac{(v_0-v_E)^2}{2g}}$ and the time for this height to be attained $\boxed{T'=\dfrac{(v_0-v_E)}{g}}$. We note that both of these are less than those obtained in the ground's frame.


Observation : The maximum height attained by the ball in the ground's frame is $H=\dfrac{v_0^2}{2g}$ at a time $T_H=\dfrac{v_0}{g}$. We note that this time is the same as that in the elevator's frame : i.e. $T'_H=T_H= \dfrac{v_0}{g}$ as a consequence of time being an invariant between two events. The distance of the ball at this height from the elevator would however be different : $y'(H)\ne H$, the events not being simultaneous. In fact, $y'(H)=H-v_Et_H=H-\dfrac{v_Ev_0}{g}$ by the Galilean transformations.

Request : I would like to ask @Hill or someone else, are my reasoning and conclusions correct? The problem is conceptually good.

 

[TSny]

Request : I would like to ask @Hill or someone else, are my reasoning and conclusions correct?

It all looks good to me.

 

[PeroK]

My solution:

Spoiler

In the ground frame:

Mary's vertical velocity is $+v$ and Bob's is $-v$. The high point relative to Mary is when the clock is moving up at $+v$ and the high point relative to Bob is when the clock is moving down at $-v$.

The time difference between these two is $0.07s$. Due to the symmetry of projectile motion (we've been through this before!), the clock takes $0.035s$ to accelerate from $0$ to $v$. With $g$ as specified, this gives $v =0.35m/s$.

The elevator is moving 30 degrees to the horizontal, so the speed of the elevator is twice the vertical speed. ($v = v_E\sin \theta$).

That gives $v_E = 0.7m/s$.

 

[brotherbobby]

The high point relative to Mary is when the clock is moving up at +v and the high point relative to Bob is when the clock is moving down at −v.

I'd like you to explain this bit.
At the highest point, which is different for Bob and Mary, the clock is at rest relative to either of them.

 

[PeroK]

I'd like you to explain this bit.
At the highest point, which is different for Bob and Mary, the clock is at rest relative to either of them.

At the highest point of projectile motion, the vertical component of velocity is zero.

 

[brotherbobby]

Here's my attempt

Problem statement :

Attempt : Bob is travelling down the blue escalator with a speed $v_E$, and Mary up in the green escalator with the same speed. At time $t=0$ Bob throws a clock up with a velocity $v_0$ relative to him. Mary, moving at $2v_E\sin\theta$ (vertically) relative to Bob, would see the clock move at speed $v'_0=v_0-2v_E\sin\theta$. If the time to reach maximum height for Mary is some $T'$, then $T'=\frac{v'_0}{g}=\frac{v_0-2v_E\sin\theta}{g}$. But $T=\frac{v_0}{g}$, the time the clock takes to attain maximum height for Bob. Hence, we have $T'=T-\frac{2v_E\sin\theta}{g}\Rightarrow \frac{2v_E\sin\theta}{g}=T-T'=\Delta T = 0.07\;\text{s}\quad\text{(given)}.$ Rearranging, we have $E = \frac{g\Delta T}{2\sin\theta}=\frac{10\times 0.07}{2\sin 30^{\circ}}=\boxed{0.7\,\text{m/s}}$.

 

[brotherbobby]

At the highest point of projectile motion, the vertical component of velocity is zero.

Correct, so when you wrote,

The high point relative to Mary is when the clock is moving up at +v

you meant $+v$ relative to the ground, right?

 

[PeroK]

Here's a shorter, trickier solution.

Note that Bob must throw the clock up at a minimum of $0.7m/s$. Otherwise, the whole experiment lasts less than $0.07s$.

Consider the time when the clock is moving up at this speed relative to Bob. It has $0.07$ seconds until it reaches the highest point relative to Bob. So, it must be at the highest point relative to Mary at this time. Therefore, Mary is moving vertically at $0.7m/s$ relative to Bob. And $v = 0.35m/s$. Etc.

 

[PeroK]

Correct, so when you wrote,

you meant $+v$ relative to the ground, right?

I did it all in the ground frame! More or less.