A body is projected horizontally with a velocity - 23 ms -1 from a height of 5 m- What is the velocity of the body on reaching the ground-

The correct option is B 11 ms ^ 1 Total energy is conserved.Initial energy = 1 2 m× 23+m× 9.8× 5 =11.5m+49m =60.5m Fnal energy = 1 ...

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Standard XII

Physics

Pure Rolling

A body is pro...

Question

A body is projected horizontally with a velocity $\sqrt{23} {m s}^{- 1}$ from a height of $5$ m. What is the velocity of the body on reaching the ground?

22 {m s}^{- 1}

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11 {m s}^{- 1}

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62 {m s}^{- 1}

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12 {m s}^{- 1}

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Solution

The correct option is B $11 {m s}^{- 1}$
Total energy is conserved.
Initial energy $= \frac{1}{2} m \times 23 + m \times 9.8 \times 5 = 11.5 m + 49 m = 60.5 m$
Fnal energy $= \frac{1}{2} m v^{2} \frac{1}{2} m v^{2} = 60.5 m v^{2} = 121 m^{2} / s^{2} v = 11 m / s$

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A body is projected horizontally with a velocity √23 ms−1 from a height of 5 m. What is the velocity of the body on reaching the ground?

The correct option is B 11 ms−1Total energy is conserved.Initial energy =12m×23+m×9.8×5=11.5m+49m=60.5mFnal energy =12mv212mv2=60.5mv2=121m2/s2v=11m/s

A body is projected horizontally with a velocity $\sqrt{23} {m s}^{- 1}$ from a height of $5$ m. What is the velocity of the body on reaching the ground?

The correct option is B $11 {m s}^{- 1}$
Total energy is conserved.
Initial energy $= \frac{1}{2} m \times 23 + m \times 9.8 \times 5 = 11.5 m + 49 m = 60.5 m$
Fnal energy $= \frac{1}{2} m v^{2} \frac{1}{2} m v^{2} = 60.5 m v^{2} = 121 m^{2} / s^{2} v = 11 m / s$