Before You Begin

Question #42408

Difficulty: 8/10 | 🟢 Free

The Question

A particle of mass $m$ is projected vertically upwards from the ground with an initial velocity $u = v_t$ , where $v_t = \sqrt{mg/k}$ is the terminal velocity of the particle in the medium. The medium exerts a drag force $F_d = -k v^2$ , where $v$ is the instantaneous velocity and $k$ is a positive constant. What is the total time of flight of the particle before it returns to the ground?

Options

A.
$\frac{v_t}{g} \left( \frac{\pi}{4} + \ln(\sqrt{2}+1) \right)$
B.
$\frac{v_t}{g} \left( \frac{\pi}{4} + \ln(\sqrt{2}) \right)$
C.
$\frac{v_t}{g} \left( \frac{\pi}{2} + \ln(\sqrt{2}+1) \right)$
D.
$\frac{v_t}{g} \left( \frac{\pi}{4} + \frac{1}{2}\ln(3) \right)$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Time of Flight Logic|Power-Velocity Relationship|Variable Force Integration

Why This Question Is Tricky

Common Mistakes

Students often incorrectly assume that the time of ascent equals the time of descent, a symmetry that only holds in a vacuum.

The inclusion of a drag force proportional to $v^2$ breaks this symmetry, requiring the setup of separate non-linear differential equations for ascent and descent.

The trap lies in failing to recognize that energy dissipation couples the vertical velocity decay to the terminal velocity $v_t$ , necessitating the calculation of the exact return velocity $v_f$ via spatial integration of the ODE before finding the descent time.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Motion in a Plane - Class 11 Part 1
Concept	Time of Flight Logic
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free