A) \[0\]
B) D
C) 2D
D) 4D
Correct Answer: C
Solution :
The displacement of the body during the time as it reaches the point of projection. \[\Rightarrow \] \[s=0\] \[\Rightarrow \] \[{{V}_{0}}t-\frac{1}{2}g{{t}^{2}}=0\] \[\Rightarrow \] \[t=\frac{2{{V}_{0}}}{9}\] During the same time t, the body moves is absence of gravity through a distance \[D=\upsilon t\]. because in absense of gravity \[g=0\] \[\Rightarrow \] \[D'={{V}_{0}}\left( \frac{2{{V}_{0}}}{g} \right)=\frac{2V_{0}^{2}}{g}\] In presence of gravity, the total distance covered is \[D=2H=2\frac{V_{0}^{2}}{2g}=\frac{{{V}^{2}}0}{g}\]
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