A) \[\frac{5}{4g}v_{0}^{2}\]
B) \[\frac{4}{5g}v_{0}^{2}\]
C) \[\frac{4}{3}v_{0}^{2}\]
D) \[\frac{3}{5}v_{0}^{2}\]
Correct Answer: B
Solution :
Key Idea: If u be the initial velocity and a is the projectile motion, then range \[=\frac{{{u}^{2}}sin\,2\alpha }{g}.\] Let \[\alpha \]be the angle of projection, then range = 2 (greatest height attained) \[\Rightarrow \] \[\frac{v_{0}^{2}\sin 2\,\alpha }{g}=2\left( \frac{v_{0}^{2}{{\sin }^{2}}\alpha }{2g} \right)\] \[\Rightarrow \]\[2\sin \alpha \cos \alpha ={{\sin }^{2}}\alpha \] \[\Rightarrow \]\[\tan \alpha =2\] ?(i) \[\therefore \]Range\[=\frac{v_{0}^{2}}{g}\sin 2\alpha \] \[=\frac{v_{0}^{2}}{g}\left( \frac{2\tan \alpha }{1+{{\tan }^{2}}\alpha } \right)\] \[=\frac{v_{0}^{2}}{g}\left( \frac{2.2}{1+{{2}^{2}}} \right)\] [from (i)] \[=\frac{4v_{0}^{2}}{5g}\]
You need to login to perform this action.
You will be redirected in
3 sec
