ExamBro
ExamBro
JEE Mains · Physics · STD 11 - 2. motion in straight line

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 \(g =10 \; m / s ^{2}\)).

  1. A \(1\)
  2. B \(3\)
  3. C \(5\)
  4. D \(7\)
Verified Solution

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 \(=10-5=5 \,m\)
Same subject
Explore more questions on app