A) 100 m
B) 200 m
C) 400 m
D) 800 m
Correct Answer: A
Solution :
Key Idea : In order to obtain maximum range the body should be projected at an angle of \[{{45}^{o}}.\] Horizontal range = horizontal velocity \[\,\times \] time H= Height R = Range \[R={{u}_{x}}\times T\] \[R=(u\,\cos \,\theta )\times \frac{2u\sin \theta }{g}\] \[R=\frac{{{u}^{2}}2\sin \theta \cos \theta }{g}\] \[R=\frac{{{u}^{2}}\sin 2\theta }{g}\] For maximum horizontal range, \[2\sin 2\theta =1\] \[\therefore \] \[\theta ={{45}^{o}}\] \[\therefore \] \[400=\frac{{{u}^{2}}}{g}\] ?(i) Also, maximum height of projectile is\[H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}=\frac{{{u}^{2}}{{\sin }^{2}}{{45}^{o}}}{2g}=\frac{400}{2}\times \frac{1}{2}=100m\]You need to login to perform this action.
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