A) \[\sqrt{\frac{2h}{g}}\]
B) \[\sqrt{\frac{2l}{g}}\]
C) \[\frac{1}{\sin \theta }\sqrt{\frac{2h}{g}}\]
D) \[\sin \theta \sqrt{\frac{2h}{g}}\]
Correct Answer: C
Solution :
Force down the plane\[=mg\sin \theta \] \[\therefore \] Acceleration down the plane \[=g\sin \theta \] Since \[l=0+\frac{1}{2}g\sin \theta {{t}^{2}}\] \[\therefore \]\[{{t}^{2}}=\frac{2l}{g\sin \theta }=\frac{2h}{g{{\sin }^{2}}\theta }\Rightarrow t=\frac{1}{\sin \theta }\sqrt{\frac{2h}{g}}\]
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