JEE MAIN - Physics (2022 - 24th June Evening Shift - No. 11)
An object of mass 5 kg is thrown vertically upwards from the ground. The air resistance produces a constant retarding force of 10 N throughout the motion. The ratio of time of ascent to the time of descent will be equal to : [Use g = 10 ms$$-$$2].
1 : 1
$$\sqrt 2 $$ : $$\sqrt 3 $$
$$\sqrt 3 $$ : $$\sqrt 2 $$
2 : 3
Explanation
Let time taken to ascent is t1 and that to descent is t2. Height will be same so
$$H = {1 \over 2} \times 12t_1^2 = {1 \over 2}\times8t_2^2$$
$$ \Rightarrow {{{t_1}} \over {{t_1}}} = {{\sqrt 2 } \over {\sqrt 3 }}$$
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