JEE Mains · Physics · STD 11 - 2. motion in straight line
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 \(\left( g =10 \; ms ^{-2}\right)\).
- A \(6\)
- B \(5\)
- C \(4\)
- D \(3\)
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 \(h _{1}=50 t -\frac{1}{2} gt ^{2}\) \(h _{2}=50( t -2)-\frac{1}{2} g ( t -2)^{2}\) \(h _{1}= h _{2}\)…
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