question_answer

The horizontal range of a projectile is\[400m\]. The maximum height attained by it will be:

A) \[100m\]                             

B) \[200m\]

C) \[400m\]                             

D) \[800m\]

Correct Answer: A

Solution :

The horizontal range = horizontal velocity                                                                           \[\times \]time                 \[R={{u}_{x}}\times T\]                 \[R=(u\cos \theta )\times \frac{2u\sin \theta }{g}\]                 \[R=\frac{{{u}^{2}}2\sin \theta \cos \theta }{g}\]                 \[R=\frac{{{u}^{2}}\sin 2\theta }{g}\] For maximum horizontal range\[\sin 2\theta =1\] \[\therefore \]  \[\theta ={{45}^{o}}\] \[\therefore \]  \[400=\frac{{{u}^{2}}}{g}\]                                           ... (i) Also, maximum height of projectile \[H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}=\frac{{{u}^{2}}{{\sin }^{2}}}{2g}\frac{{{45}^{o}}}{{}}=\frac{400}{2}\times \frac{1}{2}=100m\]