Home
Class 11
PHYSICS
A body is allowed to fall from a height ...

A body is allowed to fall from a height of 100 m. If the time taken for the first 50 m is `t_(1)` and for the remaining 50 m is `t_(2)`
The ratio of `t_(1)` and `t_(2)` is nearly

A

`5:2`

B

`3:1`

C

`3:2`

D

`5:3`

Text Solution

Verified by Experts

The correct Answer is:
A
Promotional Banner

Topper's Solved these Questions

  • KINEMATICS

    AAKASH SERIES|Exercise ADDITIONAL PRACTICE EXERCISE (LEVEL-I I) (LECTURE SHEET (ADVANCED)) (Matrix Matching Type Questions)|1 Videos
  • KINEMATICS

    AAKASH SERIES|Exercise ADDITIONAL PRACTICE EXERCISE (LEVEL-I I) (LECTURE SHEET (ADVANCED)) (Integer Type Questions)|1 Videos
  • KINEMATICS

    AAKASH SERIES|Exercise ADDITIONAL PRACTICE EXERCISE (LEVEL-I I) (LECTURE SHEET (ADVANCED)) (More than One correct answer Type Questions)|3 Videos
  • HORIZONTAL CIRCULAR MOTION

    AAKASH SERIES|Exercise ADDITIONAL PRACTICE EXERCISE (PRACTICE SHEET (ADVANCED) Integer Type Questions)|3 Videos
  • KINETIC THEORY OF GASES

    AAKASH SERIES|Exercise Practice Exercise|15 Videos

Similar Questions

Explore conceptually related problems

A particle encute SHM between x = -A and x = +A , The time taken for it to go from 0 to A//2 is T_(1) and to go from A//2 to A is T_(2) , then

A vehicle starts moving in a straight line with an acceleration, a = 4 m//s^(2) , with initial velocity equal to zero. After accelerating for time t_(1) , the vehicle moves uniformly and for time t_(2) , the vehicle finally decelerates for time, t_(1) eventually coming to a stop. The total time taken during the motion is 10s and the average velocity during the motion is 5.1 m/s. The time taken by the vehicle during acceleration is

An organic compound undergoes first order decomposition. The time taken for its decomposition to 1/8 and 1/10 of its initial concentration are t_(1/8) and t_(1//10) respectively. What is the value of ([t_(1//8)])/([t_(1//10)])xx10 ? (log_(10)2=0.3)