Two balls are dropped from the same heig...

Two balls are dropped from the same height at two different places A and B where the acceleration due to gravities are `g_A and g_B`. The body at 'B' takes 't' seconds less to reach the ground and strikes the ground with a velocity greater than at 'A' by `upsilon m//s`. Then the value of `upsilon//t` is

`(1)/sqrt(g_(A)g_(B))`

`2sqrt(g_(A)g_(B))`

`(1)/(g_(A)g_(B))`

`sqrt(g_(A)g_(B))`

Text Solution

Verified by Experts

The correct Answer is:

Similar Questions

Explore conceptually related problems

Two bodies whose masses are in the ratio 2:1 are dropped simultaneously at two places A and B where the accelerations due to gravity are g_A and g_B respectively. If they reach the ground simultaneously, the ratio of the heights from which they are dropped is

A body released from the top of a tower of height h takes T seconds to reach the ground. The position of the body at T/4 seconds is

An object falling freely from rest reaches ground in 2 s. If acceleration due to gravity is 9.8 "m s"^(-2) , then what will be the velocity of object on reaching the ground?

A body is released form the top of a tower of height h meters. It takes t seconds to reach the ground. Where is the ball at the time t/2 sec ?

The displacement of a body is given by s=(1)/(2)g t^(2) where g is acceleration due to gravity. The velocity of the body at any time t is

A ball is released from the top of a tower of height h metre. It takes T second to reach the ground. What is the position of the ball in T/3 second?

Text Solution

Similar Questions

Recommended Questions