1. Given data

• Mass of object: $m = 5\,\text{kg}$
• Gravitational acceleration: $g = 10\,\text{m s}^{-2}$
• Air resistance: constant retarding force $F_r = 10\,\text{N}$

Since air resistance always opposes motion:

• During ascent, it acts downward.
• During descent, it acts upward.

2. Acceleration during ascent

During upward motion, both gravity and air resistance act downward.

Total downward force: $mg + 10 = 5\cdot 10 + 10 = 60\,\text{N}$

So downward acceleration during ascent is $a_1 = \frac{60}{5} = 12\,\text{m s}^{-2}$

Let initial speed be $u$. At the highest point, final velocity $v=0$. Using $v = u - a_1 t_1$ we get $0 = u - 12 t_1$ $t_1 = \frac{u}{12}$

3. Maximum height reached

Using $v^2 = u^2 - 2a_1 h$ $0 = u^2 - 2(12)h$ $h = \frac{u^2}{24}$

4. Acceleration during descent

During downward motion, gravity acts downward and air resistance upward.

Net downward force: $mg - 10 = 50 - 10 = 40\,\text{N}$

So downward acceleration during descent is $a_2 = \frac{40}{5} = 8\,\text{m s}^{-2}$

The object starts descending from rest from height $h = \frac{u^2}{24}$.

Using $h = \frac{1}{2} a_2 t_2^2$ $\frac{u^2}{24} = \frac{1}{2}(8)t_2^2 = 4t_2^2$ $t_2^2 = \frac{u^2}{96}$ $t_2 = \frac{u}{4\sqrt{6}}$

5. Ratio of time of ascent to time of descent

$\frac{t_1}{t_2} = \frac{u/12}{u/(4\sqrt{6})} = \frac{4\sqrt{6}}{12} = \frac{\sqrt{6}}{3}$

Now, $\frac{\sqrt{2}}{\sqrt{3}} = \sqrt{\frac{2}{3}} = \frac{\sqrt{6}}{3}$

Hence, $t_1 : t_2 = \sqrt{2} : \sqrt{3}$

6. Checking options

• A: $1:1$ ❌
• B: $\sqrt{2}:\sqrt{3}$ ✅
• C: $\sqrt{3}:\sqrt{2}$ ❌
• D: $2:3$ ❌

Therefore, the correct option is B.