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Theoretical Verification of Law of Conservation

A body is pro...

Question

A body is projected vertically up with velocity ${98 m s}^{- 1}$ . After $2 s$ if the acceleration due to gravity of earth disappears, the velocity of the body at the end of next $s e c o n d$ is:

{49 m s}^{- 1}

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{49.6 m s}^{- 1}

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{78.4 m s}^{- 1}

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{94.7 m s}^{- 1}

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Solution

The correct option is C ${78.4 m s}^{- 1}$
Velocity after $2 s e c$ will be $v = u - g t = 98 - 9.8 (2) = 78.4 m / s$
As after $t = 2 s e c$ the retardation due to gravity disappear so the velocity will $n o t$ change $f u r t h e r$ so velocity
at $t = 3 s e c$ will also be $78.4 m / s$

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A body is projected vertically up with velocity 98ms−1. After 2s if the acceleration due to gravity of earth disappears, the velocity of the body at the end of next second is:

The correct option is C 78.4 ms−1Velocity after 2sec will be v=u−gt=98−9.8(2)=78.4m/sAs after t=2sec the retardation due to gravity disappear so the velocity will not change further so velocity at t=3sec will also be 78.4m/s

A body is projected vertically up with velocity ${98 m s}^{- 1}$ . After $2 s$ if the acceleration due to gravity of earth disappears, the velocity of the body at the end of next $s e c o n d$ is:

The correct option is C ${78.4 m s}^{- 1}$
Velocity after $2 s e c$ will be $v = u - g t = 98 - 9.8 (2) = 78.4 m / s$
As after $t = 2 s e c$ the retardation due to gravity disappear so the velocity will $n o t$ change $f u r t h e r$ so velocity
at $t = 3 s e c$ will also be $78.4 m / s$