question_answer

A body falling for 2s covers a distance S equals to that covered in next second. Find the value of S.       \[(g=10\text{ }m/{{s}^{2}})\]

Answer:

Distance travelled by the falling body in first two sec is \[S=2u+\frac{1}{2}g{{(2)}^{2}}=2u+2g=2(u+g)\to (1)\] Distance travelled by the same body in the third second is \[{{S}_{1}}=u+g\left( 3-\frac{1}{2} \right)=u+\frac{5g}{2}\to (2)\] Given\[S={{S}_{1}}\] \[2(u+g)=u+\frac{5\times g}{2}\]on solving\[u=5m{{s}^{-1}}\] From eq (1) \[S=2(u+g)=2(5+10)=30m\]