Kinematics Question 417
Question: A body is thrown upwards. If air resistance causing deceleration of $ 5m/s^{2} $ , then ratio of time of ascent to time of descent it’s $ [takeg=10m/s^{2}] $
Options:
A) $ \sqrt{\frac{1}{2}} $
B) $ \sqrt{\frac{1}{2.5}} $
C) $ \sqrt{\frac{1}{3}} $
D) $ \sqrt{\frac{1}{5}} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \frac{~\text{ Time of ascent}}{\text{Time of descent}}=\frac{( \frac{u}{g+a} )}{\frac{u}{\sqrt{( g+a )( g-a )}}} $
$ =\frac{\sqrt{( g+a )( g-a )}}{g+a}=\sqrt{\frac{( g-a )}{g+a}} $
$ =\sqrt{\frac{10-5}{10+5}}=\sqrt{\frac{5}{15}}=\sqrt{\frac{1}{3}} $
BETA
