JEE Mains · Physics · STD 11 - 2. motion in straight line
From a tower of height \(H\), a particle is thrown vertically upwards with a speed \( u\). The time taken by the particle, to hit the ground, is \(n\) times that taken by it to reach the highest point of its path. The relation between \(H, u\) and \(n\) is:
- A \(gH=(n-2)^2u^2\)
- B \(2gH=nu^2(n-2)\)
- C \(gH=(n-2)u^2\)
- D \(2gH=n^2u^2\)
Answer & Solution
Correct Answer
(B) \(2gH=nu^2(n-2)\)
Step-by-step Solution
Detailed explanation
\begin{array}{l} Speed\,on\,reaching\,ground\,v = \sqrt {{u^2} + 2gh} \\ Now,\,v = u + at\\ \Rightarrow \,\,\,\,\sqrt {{u^2} + 2gh} = - u + gt\\ Time\,taken\,to\,reach\,highest\,{\rm{point}}\,is\,t = \frac{u}{g},\\ \Rightarrow t = \frac{{u + \sqrt {{u^2} + 2gH} }}{g} =…
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