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 \[H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}=\frac{{{u}^{2}}{{\sin }^{2}}}{2g}\frac{{{45}^{o}}}{{}}=\frac{400}{2}\times \frac{1}{2}=100m\]You need to login to perform this action.
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