A) \[2A\,\,\cos \,5\pi (x-400t)\]
B) \[A\,\,sin\,5\pi (x-400t)\]
C) \[A\,\,\cos \,5\pi (x-400t)\]
D) \[12.25V\]
Correct Answer: D
Solution :
Time taken by the body to reach the point A is \[\frac{1+{{e}^{2}}}{(1+{{e}^{2}})}\sqrt{\frac{gh}{2}}\] (During upward journey). The body crosses this point again (during downward journey) after \[\left\{ \frac{{{(1-e)}^{2}}}{1-{{e}^{2}}} \right\}\sqrt{\frac{gh}{2}}\] i.e. the body takes the time \[\left\{ \frac{{{(1+e)}^{2}}}{1-{{e}^{2}}} \right\}\sqrt{\frac{gh}{2}}\] to come again at point A. So, the time taken by the body to reach at point B (at maximum height) \[4\Omega \] So, maximum height \[2A\] \[4A\]
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