Answer:
Explanation:
Given
When we drop an object from height , suppose h
it takes time T
using equation of motion
[tex]h=ut+\frac{1}{2}at^2[/tex]
where
[tex]h=displacement[/tex]
[tex]u=initial\ velocity[/tex]
[tex]a=acceleration[/tex]
[tex]t=time[/tex]
here [tex]u=0[/tex] because it dropped from a certain height
[tex]h=\frac{1}{2}gT^2[/tex]
[tex]T=\sqrt{\frac{2h}{g}}[/tex]
When height is increases to three times of original height
i.e. [tex]h'=3 h[/tex]
then time period becomes
[tex]T'=\sqrt{\frac{2\times 3h}{g}}[/tex]
[tex]T'=\sqrt{3}T[/tex]
