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

A ball is projected vertically upward with an initial velocity of \(50 \; ms ^{-1}\) at \(t =0 \; s\). At \(t =2 \,s\). another ball is projected vertically upward with same velocity. At \(t= \; \dots \; s\), second ball will meet the first ball \(\left( g =10 \; ms ^{-2}\right)\).

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Answer & Solution

Correct Answer

(A) \(6\)

Step-by-step Solution

Detailed explanation

Let they meet at \(t = t\)

So first ball gets \(t \; sec\).

and \(2^{\text {nd }}\) gets \(( t -2) \; sec\) . and they will meet at same height

\(h _{1}=50 t -\frac{1}{2} gt ^{2}\)

\(h _{2}=50( t -2)-\frac{1}{2} g ( t -2)^{2}\)

\(h _{1}= h _{2}\)

\(50 t -\frac{1}{2} gt ^{2}=50( t -2)-\frac{1}{2} g ( t -2)^{2}\)

\(100=\frac{1}{2} g \left[ t ^{2}-( t -2)^{2}\right]\)

\(100=\frac{10}{2}[4 t -4]\)

\(5= t -1\)

\(t =6 \; sec\)

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