question_answer

The maximum height of a projectile is half of its range on the horizontal. If the velocity of projection is u, its range on the horizontal is

A)  \[\frac{2{{u}^{2}}}{5g}\]              

B)         \[\frac{3{{u}^{2}}}{5g}\]

C)  \[\frac{{{u}^{2}}}{g}\]                  

D)         \[\frac{{{u}^{2}}}{5g}\]

E)  \[\frac{4{{u}^{2}}}{5g}\]

Correct Answer: E

Solution :

Maximum height \[=Range\times \frac{1}{2}\] \[\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}=\frac{{{u}^{2}}\sin 2\theta }{g}\times \frac{1}{2}\] \[\frac{{{\sin }^{2}}\theta }{2}=\frac{2\sin \theta .\cos \theta }{2}\]                 \[\tan \theta =2\] \[\therefore \]  \[\sin \theta =\frac{2}{\sqrt{5}}and\,\cos \theta =\frac{1}{\sqrt{5}}\] \[\therefore \]  \[R=\frac{2{{u}^{2}}\sin \theta .\cos \theta }{g}\]                 \[\frac{2{{u}^{2}}\times \frac{2}{\sqrt{5}}\times \frac{1}{\sqrt{5}}}{g}\]                 \[R=\frac{4{{u}^{2}}}{5g}\]