A) h/4
B) h/2
C) 7h/8
D) 3h/4
Correct Answer: D
Solution :
Suppose ball is thrown upwards with a initial velocity u, then \[I=0.5\] \[I=0.5\] (\[\frac{1}{2}iBL\]\[{{\tan }^{-1}}\left( \frac{3}{4} \right)\]) \[{{\tan }^{-1}}\left( \frac{4}{5} \right)\] \[{{\tan }^{-1}}\left( \frac{4}{3} \right)\] ??(i) Now from second equation of motion \[{{\sin }^{-1}}\left( \frac{3}{5} \right)\] \[I={{I}_{1}}\cos \,\omega t+{{I}_{2}}\sin \omega t\] \[\frac{1}{\sqrt{2}}({{I}_{1}}+{{I}_{2}})\] [from Eq. (i)] \[\frac{1}{\sqrt{2}}{{({{I}_{1}}+{{I}_{2}})}^{2}}\] \[\frac{1}{\sqrt{2}}{{(I_{1}^{2}+I_{2}^{2})}^{1/2}}\] So, \[\frac{1}{2}{{(I_{1}^{2}+I_{2}^{2})}^{1/2}}\] .....(ii) Again from second equation of motion after putting \[f\] \[f\] \[f\] [from Eq.(i)] \[f<{{f}_{r}}\] [from Eq.(ii)] \[f={{f}_{r}}\]
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