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A ball of mass0.5 \mathrm{~kg} is dropped from the height of 10 \mathrm{~m}. The height, at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity, is___________ \mathrm{m}. [Use g=10 \mathrm{~m} / \mathrm{s}^{2} ]

Option: 1

5


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{m= 0.5\, kg}
\mathrm{h= 10\, m}
\mathrm{v= 10\left ( \frac{m}{s} \right )}


\mathrm{ \sqrt{2gy}= 10}
\mathrm{ 2\times 10\times y= 100}
\mathrm{ y= 5\, m}

At heights 5 m from the surface or from the top the velocity becomes equal to the magnitude of acceleration due to gravity

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rishi.raj

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