A stone is dropped from a height h simultaneously another stone is thrown up from the ground with such a velocity so that it can reach a height of 4h. The time when two stones cross each other is `sqrt(((h)/(kg)))` where `k = "_____"`

A stone is dropped from a height h simultaneously another stone is thr

1.	A stone is dropped from a height h simultaneously another stone is thrown up from the ground with such a velocity so that it can reach a height of 4h. The time when two stones cross each other is `sqrt(((h)/(kg)))` where `k = "_____"`
Answer» Correct Answer - 8 `S_(1)=(1)/(2)g t^(2), S_(2)=ut-(1)/(2)g t^(2)` `S_(1)+S_(2)=h,4h=(u^(2))/(2g)impliesu=sqrt(8gh)` `impliesut=h` `impliessqrt(8gh)t=h` `impliest=sqrt((h)/(8g))`

Discussion

No Comment Found

Related InterviewSolutions

Speeds of two identical cars are u and 4u at at specific instant. The ratio of the respective distances in which the two cars are stopped from that instant isA. `1:1`B. `1:4`C. `1:8`D. `1:16`
A train starts from rest with constant acceleration `a=1 m//s^(2)` A passenger at a distance S from the train runs at this maximum velocity of `10 m//s` to catch the train at the same moment at which the train starts. Find the speed of the train when the passenger catches it for the critical distance:A. `8 m//s`B. `10 m//s`C. `12 m//s`D. `15 m//s`
A train starts from rest with constant acceleration `a=1 m//s^(2)` A passenger at a distance S from the train runs at this maximum velocity of `10 m//s` to catch the train at the same moment at which the train starts. If `S=25.5` m and passenger keeps running, find the time in which he will catch the train:A. 5 secB. 4 secC. 3 secD. `2sqrt(2)` sec
A train starts from rest with constant acceleration `a=1 m//s^(2)` A passenger at a distance S from the train runs at this maximum velocity of `10 m//s` to catch the train at the same moment at which the train starts. Find the critical distance `S_(c)` for which passenger will take the ten seconds time to catch the trainA. 50 mB. 35 mC. 30 mD. 25 m
A particle is moving along X-axis under a force such that its position -time graph is as shown in figure. As the particle passes position (6) :A. it is moving along positive X-direction with a speed that is increasing with timeB. it is moving along positive X-direction with a speed that is decreasing with timeC. it is moving along negative X-direction with a speed that is increasing with timeD. it is moving along negative X-direction with a speed that is decreasing with time
A particle is moving along X-axis under a force such that its position -time graph is as shown in figure. As the particle passes position (5) :A. it is instantaneously at rest and will now move along positive X-directionB. it is moving along positive X-direction with a speed that is decreasing with timeC. it is moving along negative X-direction with a maximum speedD. it is moving along negative X-direction with a minimum speed
A particle is moving along X-axis under a force such that its position -time graph is as shown in figure. As the particle passes position (1):A. it is moving along negative X-direction with a speed that is increasing with timeB. it is moving along positive X-direction with a speed that is decreasing with timeC. it is moving along negative X-direction with a speed that is decreasing with timeD. it is moving along positive X-direction with a speed that is increasing with time
A particle is moving along X-axis under a force such that its position -time graph is as shown in figure. As the particle passes position (2):A. it is moving along negative X-direction with a maximum speedB. it is moving along positive X-direction with a minimum speedC. it is moving along negative X-direction and its speed is zero hereD. it is moving along negative X-direction with a minimum speed
A body is released from the top of a tower of height h. It takes t sec to reach the ground. Where will be the ball after time `(t)/(2)` sec?A. `(h)/(2)`B. `(h)/(4)`C. `(3h)/(4)`D. `(3h)/(2)`
In the question number 67, the time taken by the ball to reach the ground isA. 2 sB. 3 sC. 5 sD. 7 s

A stone is dropped from a height h simultaneously another stone is thrown up from the ground with such a velocity so that it can reach a height of 4h. The time when two stones cross each other is `sqrt(((h)/(kg)))` where `k = "_____"`

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment