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} $



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