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JEE Mains · Physics · STD 11 - Motion in straight line
  • JEE Mains 2022
  • Medium
  • Numerical

A ball of mass \(0.5 \; kg\) is dropped from the height of \(10 \; m\). The height, at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity, is \(\dots \; m\). (Use \(g =10 \; m / s ^{2}\)).

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Answer & Solution

Correct Answer

(C) \(5\)

Step-by-step Solution

Detailed explanation

\(v^{2}=u^{2}+2 a s\)

\(100=0+2(10) \,s\)

\(s=5\, m\)

Height from ground \(=10-5=5 \,m\)

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