Kinematics Question 165
A particle is thrown up inside a stationary lift of sufficient height. The time of flight is T. Now it is thrown again with same initial speed $ v _{0} $ with respect to lift. At the time of second throw, lift is moving up with speed $ v _{0} $ and uniform acceleration g upward (the acceleration due to gravity). The new time of flight is
Options:
A) $ \sqrt{2T} $
B) $ \frac{T}{2} $
C) $ T $
D) $2T$
Show Answer
Answer:
Correct Answer: B
Solution:
[b] with respect to lift initial velocity = $ v _{0} $
Acceleration = -2g Displacement = 0
$ \therefore S=ut+\frac{1}{2}at^{2} $
$ 0=v _{0}T’+\frac{1}{2}\times 2g\times T{{’}^{2}} $
$ \therefore T’=\frac{v _{0}}{g}=\frac{1}{2}\times \frac{2v _{0}}{g}=\frac{1}{2}T $
BETA
