A) \[\tan \,\theta \]
B) \[{{\tan }^{2}}\,\theta \]
C) \[\frac{2u}{g\,\sin \,\theta }\]
D) \[\frac{u}{g\,\sin \,\theta }\]
Correct Answer: B
Solution :
\[\operatorname{Potential}\,\,energy=mg\,\,{{h}_{max}}\] \[=\,\,mg\left( \frac{{{\mu }^{2}}{{\sin }^{2}}\theta }{2g} \right)\] \[=\,\,\,\frac{m{{u}^{2}}{{\sin }^{2}}\theta }{2}\] \[Kinetic\text{ }energy=\,\,\frac{1}{2}m\,{{(u\,cos\text{ }\theta )}^{2}}\] \[\therefore \,\,\,\,\,\,\,\,\,\,\frac{PE}{KE}={{\tan }^{2}}\theta \]You need to login to perform this action.
You will be redirected in
3 sec