A) \[\frac{R}{{{k}^{2}}+1}\]
B) \[\frac{R}{{{k}^{2}}-1}\]
C) \[\frac{R}{1-{{k}^{2}}}\]
D) \[\frac{R}{k+1}\]
Correct Answer: C
Solution :
Kinetic energy = Potential energy \[\frac{1}{2}m\,{{(k{{v}_{e}})}^{2}}=\frac{mgh}{1+\frac{h}{R}}\]Þ \[\frac{1}{2}m{{k}^{2}}2gR=\frac{mgh}{1+\frac{h}{R}}\] \[\Rightarrow \,h=\frac{R{{k}^{2}}}{1-{{k}^{2}}}\] Height of Projectile from the earth's surface = h Height from the centre \[r=R+h=R+\frac{R{{k}^{2}}}{1-{{k}^{2}}}\] By solving \[r=\frac{R}{1-{{k}^{2}}}\]
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