An object of mass 5kg 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 = 10ms^-2].

Option 1 - 1 : 1

Option 2 - <math><mrow><mroot><mrow><mn>2</mn></mrow><mrow></mrow></mroot><mo>:</mo><mroot><mrow><mn>3</mn></mrow><mrow></mrow></mroot></mrow></math>

Option 3 - <math><mrow><mroot><mrow><mn>3</mn></mrow><mrow></mrow></mroot><mo>:</mo><mroot><mrow><mn>2</mn></mrow></mroot></mrow></math>

Option 4 - 2 : 3

3 Views|Posted 6 months ago

Asked by Shiksha User

1 Answer

Sort by:

Answered by

Payal Gupta | Contributor-Level 10

6 months ago

Correct Option - 2

Detailed Solution:

$a = \frac{m g + 10}{g} = \frac{5 * 10 + 10}{10}$

$\Rightarrow a = 6 m / s e c^{2}$

v = u + at

0 = u – 6t

$t_{A} = \frac{u}{6}$ (time of Ascent)

$= u * \frac{u}{6} - \frac{u}{2} * 6 * \frac{u}{6} * \frac{u}{6}$

= $\frac{u^{2}}{6} - \frac{u^{2}}{12}$

$h = \frac{u^{2}}{12}$

for time of descent

$a = \frac{50 - 10}{10}$

a = 4 m /sec²

$\Rightarrow \frac{u^{2}}{12} = 0 * t + \frac{1}{2} * 4 t^{2}$

$t_{B} = \frac{u}{\sqrt[]{24}}$

$\frac{t_{A}}{t_{B}} = \frac{u * 24}{6 * u} =$ $\sqrt[]{2}$

Upvote

Taking an Exam? Selecting a College?

Get authentic answers from experts, students and alumni that you won't find anywhere else.

On Shiksha, get access to

66K

Colleges

1.2K

Exams

6.9L

Reviews

1.8M

Answers

Learn more about...

Physics Motion in Straight Line 2025

View Exam Details

Share Your College Life Experience

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

Ask Current Students, Alumni & our Experts

Quick Actions

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Have a question related to your career & education?

See what others like you are asking & answering

An object of mass 5kg 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 = 10ms-2].

Similar Questions for you