Question

Body a of mass m is droped from height of 1 m and body b of mass 3m is droped from height 9m ratio of time taken by bodies to reach the ground

Answer 1

GiveN :

• Mass of body A = m
• Body A is dropped from 1m (h1)
• Mass of body B = 3m
• Body B is dropped from 9m (h2)

To FinD :

Ratio of Time taken by both the bodies to reach ground.

SolutioN :

Initial velocity (u) would be zero in every case,

$\implies \sf{h \: = \: ut \: + \: \dfrac{1}{2} gt^2} \\ \\ \implies \sf{h \: = \: \dfrac{gt^2}{2}} \\ \\ \implies \sf{t^2 \: = \: \dfrac{2h}{g}} \\ \\ \implies {\boxed{\sf{t \: = \: \sqrt{\dfrac{2h}{g}}}}}$

__________________________

$\underbrace{\sf{Time \: of \: Body \: A}}$

$\implies \sf{t_1 \: = \: \sqrt{\dfrac{2h_1}{g}}} \\ \\ \implies \sf{t_1 \: = \: \sqrt{\dfrac{2(1)}{10}}} \\ \\ \implies \sf{t_1 \: = \: \sqrt{\dfrac{2}{10}}} \: \: \: \: \: ...(1)$

__________________________

$\underbrace{\sf{Time \: of \: Body \: B}}$

$\implies \sf{t_2 \: = \: \sqrt{\dfrac{2h_2}{g}}} \\ \\ \implies \sf{t_2 \: = \: \sqrt{\dfrac{2(9)}{10}}} \\ \\ \implies \sf{t_2 \: = \: \sqrt{\dfrac{18}{10}}} \: \: \: \: \: \: \: ...(2)$

_______________________

$\underbrace{\sf{Ratio \: of \: time \: Intervals}}$

$\implies \sf{\dfrac{t_1}{t_2} \: = \: \dfrac{\sqrt{\dfrac{2}{10}}}{\sqrt{\dfrac{18}{10}}}} \\ \\ \implies \sf{\dfrac{t_1 ^2}{t_2 ^2} \: = \: \dfrac{\dfrac{2}{10}}{\dfrac{18}{10}}} \\ \\ \implies \sf{\dfrac{t_1 ^2}{t_2 ^2} \: = \: \dfrac{2}{18}} \\ \\ \implies \sf{\dfrac{t_1 ^2}{t_2 ^2} \: = \: \dfrac{1}{9}} \\ \\ \implies \sf{\dfrac{t_1}{t_2} \: = \: \sqrt{\dfrac{1}{9}}} \\ \\ \implies \sf{\dfrac{t_1}{t_2} \: = \: \dfrac{1}{3}} \\ \\ \underline {\sf{\therefore \: Ratios \: of \: Time \: interval \: is \: 1:3}}$

Answer 2

Given:

Body of mass m is dropped from 1 m ,

Body of mass 3m is dropped from 9 m.

To find:

Ratio of time taken to reach the ground.

Concept:

Considering acceleration due to gravity to be constant along the journey , we can simply apply the equation of kinematics to solve this problem.

Calculation:

For any object , time taken be t .

$h = ut + \frac{1}{2} g {t}^{2}$

$= > h = 0 + \frac{1}{2} g {t}^{2}$

$= > {t}^{2} = \dfrac{2h}{g}$

$= > t = \sqrt{ \dfrac{2h}{g} }$

Since g is constant , we can say that :

$\: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \huge{ \red{ \bold{t \: \: \propto \: \sqrt{h} }}}}$

Let time for 1st body be t1 , and for 2nd body be t2.

$\therefore \: \: \dfrac{t1}{t2} = \sqrt{ \dfrac{h1}{h2} }$

$= > \: \: \dfrac{t1}{t2} = \sqrt{ \dfrac{1}{9} }$

$= > \: \: \dfrac{t1}{t2} = \dfrac{1}{3}$

$= > t1 : t2 = 1 : 3$

So final ratio is :

$\boxed{ \blue{ \huge{ \bold{t1 : t2 = 1 : 3}}}}$