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.A. `t_1 = t_2`B. `t_1 = 2t_2`C. `t_1 = 3t_2`D. `t_1 = (t_2)/((sqrt(2) - 1))`

A body is dropped from a height `h`. If `t_1 and t_2` be the times in

1.	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.A. `t_1 = t_2`B. `t_1 = 2t_2`C. `t_1 = 3t_2`D. `t_1 = (t_2)/((sqrt(2) - 1))`
Answer» Correct Answer - D `t_1 = sqrt((2(h//2))/(g)) = sqrt((h)/(g)) rArr t_1 + t_2 = sqrt((2h)/(g))` `rArr t_2 = sqrt((2h)/(g)) - t_1 = (sqrt(2 ) - 1) sqrt((h)/(g))` `rArr (t_1)/(t_2) = (1)/(sqrt(2) - 1) rArr t_1 = (t_2)/(sqrt(2) -1)`.

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 body `A` starts from rest with an acceleration `a_1`. After `2` seconds, another body `B` starts from rest with an acceleration `a_2`. If they travel equal distances in the `5th` second, after the start of `A`, then the ratio `a_1 : a_2` is equal to :A. `5 : 9`B. `5 : 7`C. `9 : 5`D. `9 : 7`
If a body starts from rest, the time in which it covers a particular displacement with uniform acceleration is :A. inversely alphaortional to the square root of the displacementB. inversely alphaortional to the dispalcementC. directly alphaortional to the displacementD. directly alphaortional to the square root of the displacement
A particle starts from rest with uniform acceleration and is velocity after `n` seconds `v`. The displacement of the body in last two seconds is.A. `(2v(n + 1))/(n)`B. `(v(n + 1))/(n)`C. `(v(n - 1))/(n)`D. `(2 v(n - 1))/(n)`
A particle starts from rest with uniform acceleration and is velocity after `n` seconds `v`. The displacement of the body in last two seconds is.A. `(v(n + 1))/(n)`B. `(2 v(n + 1))/(n)`C. `(2 v(n - 1))/(n)`D. `(v(n - 1))/(n)`
A body starting from rest moving with uniform acceleration has a displacement of `16 m` in first `4` seconds and `9 m` in first `3` seconds. The acceleration of the body is :A. `1 ms^-2`B. `2 ms^-2`C. `3 ms^-2`D. `4 ms^-2`
A body is moving from rest under constant acceleration and let `S_1` be the displacement in the first `(p - 1)` sec and `S_2` be the displacement in the first `p` sec. The displacement in `(p^2 - p + 1)` sec. will beA. `S_1 + S_2`B. `S_1 S_2`C. `S_1 - S_2`D. `S_1//S_2`
A particle moves along a staight line such that its displacement at any time t is given by `s=t^3-6t^2+3t+4m`. Find the velocity when the acceleration is 0.A. `3 ms^-1`B. `-12 ms^-1`C. `42 ms^-1`D. `-9 ms^-1`
The displacement of a particle moving along x-axis is given by : `x = a + bt + ct^2` The acceleration of the particle is.A. bB. cC. `b + c`D. 2 c
If the velocity of a particle is `v = At + Bt^2`, where `A` and `B` are constant, then the distance travelled by it between `1 s` and `2 s` is :A. `(3)/(2) A + 4B`B. `3 A + 7 B`C. `(3)/(2)A + (7)/(3) B`D. `(A)/(2) + (B)/(3)`

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.A. `t_1 = t_2`B. `t_1 = 2t_2`C. `t_1 = 3t_2`D. `t_1 = (t_2)/((sqrt(2) - 1))`

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment