JEE Mains · Physics · STD 11 - 2. motion in straight line
From the top of a tower, a ball is thrown vertically upward which reaches the ground in \(6 \,s\). A second ball thrown vertically downward from the same position with the same speed reaches the ground in \(1.5\, s\). A third ball released, from the rest from the same location, will reach the ground in........ \(s\).
- A \(38\)
- B \(7\)
- C \(8\)
- D \(3\)
Answer & Solution
Correct Answer
(D) \(3\)
Step-by-step Solution
Detailed explanation
Let height of tower be \(h\) and speed of projection in first two cases be \(u\). For case-I : \(2^{\text {nd }}\) equation \(s=u t+\frac{1}{2}\) at \(^{2}\) \(h =- u (6)+\frac{1}{2}\, g (6)^{2}\) \(H =-6 u +18\,g \ldots\) \((i)\) For case-II…
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