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].

Solution

Physics sir · 4 steps

Air drag always opposes motion, so it adds to gravity going up and fights gravity coming down.

1. Drag as acceleration. The retarding force gives an extra acceleration of 10/5 = 2 m s⁻², always opposite to the velocity.
2. Two different accelerations. Going up, gravity and drag both point down, so the retardation is aᵤ = 10 + 2 = 12 m s⁻². Coming down, drag points up, so the acceleration is a_d = 10 - 2 = 8 m s⁻².
3. Same height both ways. The ascent ends at rest, so treat it in reverse as starting from rest: h = 1/2aᵤ tᵤ². The descent starts from rest: h = 1/2a_d t_d². Equate the two heights.

   tᵤ/t_d = √(a_d/aᵤ) = √(8/12) = √(2/3)

4. So. The ratio of time of ascent to time of descent is √2 : √3, option (b). The ascent is quicker because the stone slows down faster than it speeds up.

The idea to keep. With constant drag, use g + f/m going up and g - f/m coming down. Over the same height, time goes as 1/√a, so the bigger acceleration means the shorter trip.

Answer from the NTA final answer key. Solution by PrepWiser.