Home
Class 12
PHYSICS
A stone falls freely under gravity. It c...

A stone falls freely under gravity. It covered distances `h_1, h_2` and `h_3` in the first `5` seconds. The next `5` seconds and the next `5` seconds respectively. The relation between `h_1, h_2` and `h_3` is :

A

`h_1=2h_2=3h_3`

B

`h_1=h_2/3=h_3/5`

C

`h_2=3h_1` and `h_3=3h_2`

D

`h_1=h_2=h_3`

Text Solution

Verified by Experts

The correct Answer is:
A
Promotional Banner

Similar Questions

Explore conceptually related problems

A body travels for 15 second starting from rest with constant acceleration. If it travels distances S_(1), S_(2) and S_(3) in the first five seconds, second five seconds and next five seconds respectively the relation between S_(1), S_(2) and S_(3) is

In previous question, what is the relation between h_1 and h_2

A particle is falling freely under gravity. In first t second it covers distance x_1 and in the next t second, it covers distance x_2, then t is given by

A stone falls freely from rest from aheight h and it travels a distance h//2 in the last second. The time of journey is

A stone falls freely from rest from aheight h and it travels a distance 9h//25 in the last second. The value of h is

A body starts to fall freely under gravity. The distances covered by it in first, second and third second are in ratio

A particle is falling freely under gravity. In frist t secnd it covers distance d_(1) and in the next t second it covers distance d_(2) , then t is given by :

A body is dropped from a height h . If t_1 and t_2 be the times in covering first half and the next distances respectively, then the relation between t_1 and t_2 is.

The distance through which a body falls in the nth second is h. The distance through which it falls in the next second is

A stone falls freely such that the distance covered by it in the last second of its motion is equal to the distance covered by it in the first 5 seconds. It remained in air for :-