A) 14.5 m
B) 24.5 m
C) 34.5 m
D) None of these
Correct Answer: B
Solution :
As, second ball falls for 2 s \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{h}_{2}}=\,\,\frac{1}{2}g{{(2)}^{2}}\] In mean time, first ball has fallen for \[\left( 2+1 \right)=3s\] So, \[{{h}_{1}}=\,\,\frac{1}{2}g{{(3)}^{2}}\] :. Distance between two balls 2 s after release of 2nd ball is \[=\,\,{{h}_{1}}-{{h}_{2}}=\frac{1}{2}\,g\,({{3}^{2}}-{{2}^{2}})\,\,=\,\,24.5m\]
You need to login to perform this action.
You will be redirected in
3 sec
