A) \[\frac{200\times {{10}^{3}}}{999\times 200\times {{10}^{3}}}=\frac{I}{5\times {{10}^{3}}}\]
B) \[\Rightarrow \]
C) \[I=\frac{5\times {{10}^{3}}}{999}=5cm\]
D) \[\frac{m{{v}^{2}}}{r}=\frac{KZ{{e}^{2}}}{{{r}^{2}}}\]
Correct Answer: A
Solution :
Key Idea: At maximum height velocity is zero. From equation of motion we have \[\therefore \] where v is final velocity, u is initial velocity. Since ball reaches maximum height, velocity at the highest point is zero. Therefore, we have \[\frac{1}{{{v}_{0}}}=\frac{1}{{{f}_{0}}}+\frac{1}{{{u}_{0}}}\] \[{{f}_{0}}=200cm\] \[{{u}_{0}}=-2km=-2\times {{10}^{5}}cm\] \[O=50m=5\times {{10}^{3}}cm\] when \[\therefore \] then \[\frac{1}{{{v}_{0}}}=\frac{1}{200}+\frac{1}{-200\times {{10}^{3}}}\] Note: If ball has to be thrown to a greater height its initial velocity should be more than the original one.
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