A) \[\frac{R}{6}\]
B) \[\frac{R}{3}\]
C) \[\frac{2R}{3}\]
D) \[R\]
Correct Answer: B
Solution :
Maximum height attained by a projectile \[h=\frac{{{v}^{2}}R}{2gR-{{v}^{2}}}\] ?.(i) Velocity of body = half the escape velocity \[v=\frac{{{v}_{e}}}{2}\] or \[v=\frac{\sqrt{2gR}}{2}\] \[\Rightarrow \] \[{{v}^{2}}\frac{2gR}{4}\] or \[{{v}^{2}}=\left( \frac{gR}{2} \right)\] Now, putting value of \[{{v}^{2}}\] in Eq. (i), we get \[h=\frac{\frac{gR}{2}.R}{2gR-\frac{gR}{2}}\] \[=\frac{g{{R}^{2}}/2}{3gR/2}\] or \[h=\frac{R}{3}\]You need to login to perform this action.
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