- #1
brotherbobby
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- 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}
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)?