Question

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\(^{-2}\)]

• A. 1 : 1
• B. \( \sqrt{2} : \sqrt{3} \)
• C. \( \sqrt{3} : \sqrt{2} \)
• D. 2 : 3

Hint:

In the presence of air resistance, the time of descent is always greater than the time of ascent because the net acceleration while falling is smaller than the acceleration while rising.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Air resistance always acts opposite to the direction of motion. This changes the net acceleration of the object during upward (ascent) and downward (descent) motion.

Step 2: Key Formula or Approach:
Let \( m = 5\text{ kg} \), \( g = 10\text{ ms}^{-2} \), and \( f = 10\text{ N} \) (air resistance). The acceleration \( a = \frac{F_{net}}{m} \). For a distance \( H \), time \( t = \sqrt{\frac{2H}{a}} \).

Step 3: Detailed Explanation:
1. During Ascent: Both gravity (\( mg \)) and air resistance (\( f \)) act downwards. \[ a_{ascent} = \frac{mg + f}{m} = \frac{5(10) + 10}{5} = \frac{60}{5} = 12\text{ ms}^{-2} \] Time of ascent \( t_a = \sqrt{\frac{2H}{12}} \).
2. During Descent: Gravity (\( mg \)) acts downwards, but air resistance (\( f \)) acts upwards. \[ a_{descent} = \frac{mg - f}{m} = \frac{5(10) - 10}{5} = \frac{40}{5} = 8\text{ ms}^{-2} \] Time of descent \( t_d = \sqrt{\frac{2H}{8}} \).
3. Ratio: \[ \frac{t_a}{t_d} = \frac{\sqrt{2H/12}}{\sqrt{2H/8}} = \sqrt{\frac{8}{12}} = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}} \]

Step 4: Final Answer
The ratio of time of ascent to the time of descent is \( \sqrt{2} : \sqrt{3} \).