A) 1 : 1
B) 11 : 1
C) 1 : 2
D) 1 : 11
Correct Answer: B
Solution :
From \[e{{q}^{n}}\,\,v=ugt\] At max. height \[v=0\] \[O=u-g(6)\] \[u=6g\text{ }m/s\] Distance covered in 1st second \[{{h}_{1}}=ut\frac{1}{2}g{{t}^{2}}\] \[{{h}_{1}}=6g(1)\frac{1}{2}g{{(1)}^{2}}\] similarly\[=\frac{11g}{2}\] ???(1) \[{{h}_{2}}=[u(7)-\frac{1}{2}g{{(7)}^{2}}]-[4(6)-\frac{1}{2}g{{(6)}^{2}}]\] \[=u-\frac{13}{2}g=6g-\frac{13}{2}g=-\frac{g}{2}\] ?..(2) Now\[\frac{{{h}_{1}}}{{{h}_{2}}}=\frac{11g/2}{-g/2}=-11/1\] \[|{{h}_{1}}:{{h}_{2}}|=11:1\]
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