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A body is released to fall under the inf...

A body is released to fall under the influence of gravity and its equation of motion is given by `s(t)=(9t^(2)+4)sec`.Find the ratio of the instantaneous velocities of the body at the fifth second to that at the seventh second after release. A). ` 2:9` B). ` 1:2` C). `1:11` D). `5:7`

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A body is released to fall under the influence of gravity and its equation of motion is given by (t)=(9t^(2)+4)sec .Find the ratio of the instantaneous velocities of the body at the fifth second and seventh second after release: A. 1:2 B. 5:7 C. 5:3 D. 1:11

The displacement of a body at any time t after starting is given by s=15t-0.4t^2 . Find the time when the velocity of the body will be 7ms^-1 .

Knowledge Check

  • 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
    `(g t^(3))/(6)`
    B
    `(g t^(2))/(2)`
    C
    gt
    D
    `(g t)/(2)`

  • A body is released from a great height and falls freely towards the earth. Exactly one sec later another body is released. What is the distance between the two bodies 2 sec after the release of the second body ?

    A
    4.9 m
    B
    9.8 m
    C
    24.5 m
    D
    50 m

  • A body falls from rest freely under gravity with an acceleration of 9.8 m//s^(2) . Neglecting air resistance, the distance travelled by the body during the third second of its motion will be:

    A
    14.7 m
    B
    24.5 m
    C
    19.6 m
    D
    29.4 m

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