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 \(g =10 \; m / s ^{2}\)).
- A \(1\)
- B \(3\)
- C \(5\)
- D \(7\)
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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