question_answer

A body is projected vertically upwards. The times corresponding to height h while ascending and while descending are \[{{t}_{1}}\] and \[{{t}_{2}}\] respectively .Then, the velocity of projection is:

A) \[g\sqrt{{{t}_{1}}{{t}_{2}}}\]

B)               \[g\frac{{{t}_{1}}{{t}_{2}}}{{{t}_{1}}+{{t}_{2}}}\]

C) \[\frac{g\,\sqrt{{{t}_{1}}{{t}_{2}}}}{2}\]                     

D) \[\frac{g\,({{t}_{1}}{{t}_{2}})}{2}\]

Correct Answer: D

Solution :

[d] Let 'u' be the velocity of projection from point A Let OA = CD = h \[{{t}_{1}}\] = time taken from OA It is also equal to time taken from C to D and the time taken from 0 to B and B to \[C={{t}_{2}}\] Also \[T=\frac{2u}{g}\] \[\frac{2u}{g}={{t}_{1}}+{{t}_{2}}\] \[u=\frac{1}{2}g+[{{t}_{2}}+{{t}_{2}}]\]