SATHEE: Kinematics Question 394

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

$∴$ Time of rise $t_{1} = \frac{u}{g + a}$

and height reached $= \frac{u^{2}}{2 (g + a)}$

Time of fall $t_{2}$ it’s given by $\frac{1}{2} (g - a) t_{2}^{2} = \frac{u^{2}}{2 (g + a)}$

$t_{2} = \frac{u}{\sqrt{(g + a) (g - a)}} = \frac{u}{(g + a)} \sqrt{\frac{g - a}{g + a}}$

$∴ t_{2} > t_{1} because \frac{1}{g + a} < \frac{1}{g - a}$

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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:

Answer:

Solution:

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