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. [JEE MAIN 2014] |
The relation between H, u and n is: |
A) \[2gH=n{{u}^{2}}(n-2)\]
B) \[gH=(n-2){{u}^{2}}\]
C) \[2gH={{n}^{2}}{{u}^{2}}\]
D) \[gH={{(n-2)}^{2}}{{u}^{2}}\]
Correct Answer: A
Solution :
[a] Time taken to reach highest point |
For time taken to reach the ground\[{{t}_{1}}=\frac{u}{g}\] |
\[-H=ut-\frac{u}{g}g{{t}^{2}}\] \[\therefore \]\[g{{t}^{2}}-2ut-2H=0\] |
\[t=\frac{2u\pm \sqrt{4{{u}^{2}}+8gH}}{2g}\] |
\[t=\frac{+\sqrt{{{u}^{2}}+2gH}}{g}\]= (− ve sign not acceptable) |
Given \[t=n\,{{t}_{1}}\] \[\frac{u+\sqrt{{{u}^{2}}+2gH}}{g}=\frac{nu}{g}\] |
Solving \[2gH=n{{u}^{2}}(n-2)\] |
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