Work Energy And Power Question 41

Question: A body of mass 2 kg is thrown up vertically with K.E. of 490 joules. If the acceleration due to gravity is 9.8 $ m/s^{2} $ , then the height at which the K.E. of the body becomes half its original value is given by [EAMCET 1986]

Options:

A) 50 m

B) 12.5 m

C) 25 m

D) 10 m

Show Answer

Answer:

Correct Answer: B

Solution:

Let h is that height at which the kinetic energy of the body becomes half its original value i.e. half of its kinetic energy will convert into potential energy mgh = $ \frac{490}{2} $

therefore $ 2\times 9.8\times h=\frac{490}{2} $

therefore $ h=12.5m. $



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