$A$ ball is thrown upward with an initial velocity $V_0$ from the surface of the earth. The motion of the ball is affected by a drag force equal to $m\gamma v^2$ (where $m$ is the mass of the ball,$v$ is its instantaneous velocity,and $\gamma$ is a constant). The time taken by the ball to rise to its zenith is:

  • A
    $\frac{1}{\sqrt{\gamma g}} \ln \left( 1 + \sqrt{\frac{\gamma}{g}} V_0 \right)$
  • B
    $\frac{1}{\sqrt{\gamma g}} \tan^{-1} \left( \sqrt{\frac{\gamma}{g}} V_0 \right)$
  • C
    $\frac{1}{\sqrt{\gamma g}} \sin^{-1} \left( \sqrt{\frac{\gamma}{g}} V_0 \right)$
  • D
    $\frac{1}{\sqrt{2\gamma g}} \tan^{-1} \left( \sqrt{\frac{2\gamma}{g}} V_0 \right)$

Explore More

Similar Questions

$A$ ball thrown vertically upwards reaches a height of $25 \,m$ in $1 \,s$. The ratio of the total distance covered by the ball in time $t=2 \,s$ and $t=4 \,s$ is (Use $g=10 \,m/s^2$)

$A$ ball is projected upwards from a height $h$ above the surface of the earth with velocity $v$. The time at which the ball strikes the ground is

$A$ particle is projected vertically upwards such that it attains a height of $h$ after $5 \ s$ and $9 \ s$ of its motion. The speed of projection is $...... \ m/s$ $(g = 10 \ m/s^2)$.

The velocity of a body upon reaching the point from which it was projected upwards is:

When a particle is thrown vertically upwards,its velocity at one third of its maximum height is $10 \sqrt{2} \, m/s$. The maximum height attained by it is .......... $m$.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo