A) \[{{h}_{1}}={{e}^{2}}h\]
B) \[{{h}_{1}}={{e}^{4}}h\]
C) \[{{h}_{1}}=eh\]
D) \[{{h}_{1}}=\frac{h}{e}\]
Correct Answer: A
Solution :
When a body falls from a height h, it strikes the ground with a velocity\[u=\sqrt{2gh}.\] Let, it rebounds with a velocity v and rise to a height\[{{h}_{1}}.\] Therefore, \[v=\sqrt{2gh}\] \[\therefore \] \[e=\frac{v}{u}=\sqrt{\left( \frac{{{h}_{1}}}{h} \right)}\] If the successive heights to which the body rebounds again and again are \[{{h}_{2}},{{h}_{3}},....,\]then \[e=\sqrt{\frac{{{h}_{1}}}{h}}=\sqrt{\frac{{{h}_{2}}}{{{h}_{1}}}}=\sqrt{\frac{{{h}_{3}}}{{{h}_{2}}}}=....\] Clearly,\[{{h}_{1}}={{e}^{2}}h,{{h}_{2}}={{e}^{4}}h...\]and so on Similarly, after nth rebound. \[{{h}_{n}}=({{e}^{2n}})h\] Note The degree of elasticity of a collision is determined by a quantity, called coefficient of restitution or coefficient of resislience of the collision. It is defined as the ratio of relative velocity of separation after collision to the relative velocity of approach before collision. It is represented by e.
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