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° 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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