A) \[0.98\text{ }m{{s}^{-1}}\]
B) \[19.6\text{ }m{{s}^{-1}}\]
C) \[4.9\,m{{s}^{-1}}\]
D) None of these
Correct Answer: A
Solution :
The given situation can be shown as \[R\cos \alpha =\frac{m{{v}^{2}}}{r}\] \[R\sin \alpha =mg\] \[\therefore \] \[\tan \alpha =\frac{gr}{{{v}^{2}}}\] Also \[\tan \alpha =\frac{r}{h}\] \[\therefore \] \[\frac{r}{h}=\frac{gr}{{{v}^{2}}}\] \[\Rightarrow \] \[{{v}^{2}}=gh=0.98\,m{{s}^{-1}}\]You need to login to perform this action.
You will be redirected in
3 sec