A body of mass $4 k g$ is thrown upward from a height of $2 m$ with an initial speed of $12 m s^{- 1}$ . Its speed at height $3 m$ is .

10.1 m s^{- 1}

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11.1 m s^{- 1}

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12.1 m s^{- 1}

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Solution

The correct option is B $11.1 m s^{- 1}$
The total mechanical energy of a body = Potential Energy + Kinetic energy
$= m g h + \frac{1}{2} m v^{2}$ .
On substituting the values of mass, height and velocity of the shot in the formula:
Potential Energy $= 4 \times 10 \times 2$ = $80 J$
Kinetic Energy = $0.5 \times 4 \times 12^{2}$ = $288 J$
Total energy = Potential energy + Kinetic energy = $368 J$
At height = 3m,
Potential Energy = $4 \times 10 \times 3 = 120 J$
Kinetic energy = Total energy – Potential energy = $248 J$
we know that,
K.E. = $\frac{1}{2}$ m $v^{2}$
v = $\sqrt{\frac{2 \times K E}{m}}$ = $\sqrt{\frac{2 \times 248}{4}}$ = $11.1 m s^{- 1}$

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A body of mass 4 kg is thrown upward from a height of 2 m with an initial speed of 12 ms−1. Its speed at height 3 m is .

A body of mass $4 k g$ is thrown upward from a height of $2 m$ with an initial speed of $12 m s^{- 1}$ . Its speed at height $3 m$ is .