Kinematics Question 394
Question: A body is thrown vertically upwards. If air resistance it’s to be taken into account, then the time during which the body rises is [assume no air resistance close to earth]
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
$ \therefore $ Time of rise $ t _{1}=\frac{u}{g+a} $
and height reached $ \text{= }\frac{u^{2}}{2( g+a )} $
Time of fall $ {{t} _{2}} $ it’s given by $ \frac{1}{2}( g-a )t _{2}^{2}=\frac{u^{2}}{2( g+a )} $
$ t _{2}=\frac{u}{\sqrt{( g+a )( g-a )}}=\frac{u}{( g+a )}\sqrt{\frac{g-a}{g+a}} $
$ \therefore {{t} _{2}}>{{t} _{1}}\text{ because }\frac{1}{g+a}<\frac{1}{g-a} $
