Kinematics Question 181
Question: A body is thrown vertically upwards. If air resistance is to be taken into account, then the time during which the body rises is [RPET 2000; KCET 2001; DPMT 2001]
Options:
A) Equal to the time of fall
B) Less than the time of fall
C) Greater than the time of fall
D) Twice the time of fall
Show Answer
Answer:
Correct Answer: B
Solution:
Let the initial velocity of ball be u Time of rise $ t _{1}=\frac{u}{g+a} $
and height reached $ =\frac{u^{2}}{2(g+a)} $
Time of fall $ t _{2} $ is given by $ \frac{1}{2}(g-a)t _{2}^{2}=\frac{u^{2}}{2(g+a)} $
$ \Rightarrow t _{2}=\frac{u}{\sqrt{(g+a)(g-a)}}=\frac{u}{(g+a)}\sqrt{\frac{g+a}{g-a}} $
$ \therefore $ $ t _{2}>t _{1} $ because $ \frac{1}{g+a}<\frac{1}{g-a} $
