A) 4s
B) \[(4+2\sqrt{2})s\]
C) \[(2+2\sqrt{2})s\]
D) \[(4+4\sqrt{2})s\]
Correct Answer: B
Solution :
[b] Let \[{{t}_{0}}=(2+t)\] second |
\[-2a=(2a)t-\frac{1}{2}a{{t}^{2}}\] |
\[\therefore \] \[{{t}^{2}}-4t-4=0\] |
\[t=\frac{4\pm \sqrt{16+16}}{2}=2+2\sqrt{2}\] |
\[\therefore \] \[{{t}_{0}}=2+t=(4+2\sqrt{2})s\] |
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