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. $
BETA
