A) 50 m
B) 60 m
C) 20 m
D) 40 m
Correct Answer: D
Solution :
| For maximum range of projectile \[\theta \] will be \[{{45}^{o}}\] by the law of projectile motion |
| \[\therefore \] \[{{R}_{\max }}=\frac{{{u}^{2}}}{g}\] |
| Given, \[u=20\,m{{s}^{-1}}\] |
| and \[g=10\,m{{s}^{-2}}\] |
| \[{{R}_{\max }}=\frac{{{(20)}^{2}}}{10}=\frac{400}{10}\] |
| \[{{R}_{\max }}=40\,m\] |
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