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एक पत्थर मुक्त रूप से गुरूत्वाधीन गिरता है। यह पत्थर पहले पाँच $(5)$ सेंकडों में $h _{1}$ दूरी, उसे अगले $5$ संकंडों में $h _{2}$ दूरी तथा उससे अगले $5$ सेंकडों में $h _{3}$ दूरी तय करता है, तो $h _{1}, h _{2}$ तथा $h _{3}$ से संबंध है
$h_1=2h_2=3h_3$
$h_1$=$\frac{{{h_2}}}{3}$=$\frac{{{h_3}}}{5}$
$h_2=3h_1$ और $h_3=3h_2$
$h_1=h_2=h_3$
Solution
$\begin{array}{l}
\,\,\,\,\,\,\,{\rm{Distance covered}}\,by\,the\,in\,first\,5\,{\rm{second}}\,\\
\,\,\left( {i.e.t = 5\,s} \right)\,is\\
\,{h_1} = \frac{1}{2} = g{\left( 5 \right)^2} = \frac{{25}}{2}g\,\,\,\,\,\,\,…\left( i \right)\\
\,\,{\rm{Distance}}\,travelled\,by\,the\\
stone\,in\,next\,5\,\,{\rm{second}}\,\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {i.e.t = 10\,s} \right)\,is\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{h_1} + {h_2} = \frac{1}{2}g{\left( {10} \right)^2} = \frac{{100}}{2}g\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,…\left( {ii} \right)\\
{\rm{Distance}}\,travelled\,by\,the\,stone\,in\,5\,{\rm{second}}\\
\left( {i.e.t = 15\,s} \right)\,\,is\\
\,\,\,\,\,\,\,\,\,\,\,\,{h_1} + {h_2} + {h_3} = \frac{1}{2}g{\left( {15} \right)^2} = \frac{{225}}{2}g\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,…\left( {iii} \right)\\
Subtract\,\left( i \right)\,from\,\left( {ii} \right),\,we\,get\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{h_1} + {h_2}} \right) – {h_2} = \frac{{100}}{2}g – \frac{{25}}{2}g = \frac{{75}}{2}g\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{h_2} = \frac{{75}}{2}g = 3{h_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,…\left( {iv} \right)\\
Subtract\,\left( {ii} \right)\,from\,\left( {iii} \right),\,we\,get\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{h_1} + {h_2} + {h_3}} \right) – \left( {{h_2} + {h_1}} \right) = \frac{{225}}{2}g – \frac{{100}}{2}g\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{h_3} = \frac{{125}}{2}g = 5{h_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,…\left( v \right)\\
From\,\left( i \right),\left( {iv} \right)\,and\,\left( v \right),\,we\,get\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{h_1} = \frac{{{h_2}}}{3} = \frac{{{h_3}}}{5}\\
\,\,\,\,\,\,\,\,
\end{array}$
