JEE Main On 8 April 2017 Question 8
Question: An object is dropped from a height h from the ground. Every time it hits the ground it looses 50% of its kinetic energy. The total distance covered as $ t\to \infty $ is - [JEE Online 08-04-2017]
Options:
A) 3h
B) $ \infty $
C) $ \frac{5}{3}h $
D) $ \frac{8}{3}h $
Show Answer
Answer:
Correct Answer: A
Solution:
- $ \frac{1}{2}mv{{’}^{2}}=\frac{1}{2}\frac{1}{2}mv^{2} $ $ v’=\frac{v}{\sqrt{2}} $ $ v=eu $ $ e=\frac{1}{\sqrt{2}} $ $ H=\lambda ( \frac{1+e^{2}}{1-e^{2}} ) $ $ =h( \frac{1+\frac{1}{2}}{1-\frac{1}{2}} )=3h $
