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 \(\left( g =10 \; ms ^{-2}\right)\).
Pick an option to check your answer.
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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