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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: 1
2. √2: √3
3. √3: √2
4. 2: 3

Correct answer: √2: √3

Solution

Ascent deceleration = (mg + f)/m = (50+10)/5 = 12 m/s^2; descent acceleration = (mg - f)/m = (50-10)/5 = 8 m/s^2. For the same height, t_ascent/t_descent = sqrt(a_descent/a_ascent) = sqrt(8/12) = sqrt(2):sqrt(3).

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