question_answer

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) \[2gH={{n}^{2}}{{u}^{2}}\]            

B) \[gH={{(n-2)}^{2}}{{u}^{2}}d\]

C) \[2gH=n{{u}^{2}}(n-2)\]         

D) \[gH=(n-2){{u}^{2}}\]

Correct Answer: C

Solution :

Speed on reaching ground \[v=\sqrt{{{u}^{2}}+2gh}\] Now, \[v=v+at\] \[\Rightarrow \,\,\sqrt{{{u}^{2}}+2gh}=-u+gt\] Time taken to reach highest point is \[t=\frac{u}{g},\] \[\Rightarrow \,\,t=\frac{u+\sqrt{{{u}^{2}}+2gH}}{g}=\frac{nu}{g}\](from question) \[\Rightarrow \,\,2gH=n(n-2){{u}^{2}}\]