Answer:
Let ABCD be the carom board of\[4ft\times 4ft\]dimensions. ‘O’ is the centre where initially Queen is placed.
Let E be the point on front edge where the Queen strikes and rebounds to fall in hole at A.
Displacement of the Queen,
(i) Displacement from centre to hole (OA) = ?
From figure, \[OA=\frac{AC}{2}\]
\[AC=\sqrt{A{{B}^{2}}+B{{C}^{2}}}=\sqrt{{{4}^{2}}+{{4}^{2}}}\,\,or\,\,4\sqrt{2}\]c
\[\Rightarrow OA=2\sqrt{2}ft\]
(ii) from centre to front edge = OE = ?
Let F be the midpoint of CD. Join of (2ft). Let FE be x. or \[DE=(2-x)\]
we can find OE, only if we know \[x.\]
In right angled \[\Delta OEF,\]
\[OE=\sqrt{{{2}^{2}}+{{x}^{2}}}\] ….(1)
Lets try to find x.
Let \[\angle EOF\,\,to\,\,\theta \]
Draw a \[\bot EG\] at E such that
\[\angle OEF=\angle GEO=\theta \]
\[20=u+\frac{a}{2}(2-1)\] \[(\because n=1\,s)\] \[20=u+\frac{a}{2}\]Alternate angles)
In \[\Rightarrow \angle EAD,tan\theta =\frac{x}{2}\] ………..(1)
\[\Rightarrow \angle AED,tan\theta =\frac{2-x}{4}\] …………..(2)
From (1) and (2)
\[\frac{x}{2}=\frac{2-x}{4}\,\,or\,\,6x=4\,\,or\,\,x=\frac{2}{3}\]
Substituting the x value in (1), we get
\[OE=\sqrt{{{2}^{2}}+{{\left( \frac{2}{3} \right)}^{2}}}\]
\[\Rightarrow \sqrt{\frac{40}{9}}=\frac{2}{3}\sqrt{10}ft\]
(iii) Displacement from front edge to hole (EA) = ? from \[\Delta AED\]
\[EA\sqrt{{{(2-x)}^{2}}+{{4}^{2}}}\]
\[=\sqrt{{{(2-\frac{2}{3})}^{2}}+{{4}^{2}}}\] \[\left( \because x=\frac{2}{3} \right)\]
\[=\sqrt{{{\left( \frac{4}{3} \right)}^{2}}+{{4}^{2}}}=\frac{4}{3}\sqrt{10}ft\]
You need to login to perform this action.
You will be redirected in
3 sec
