Question

Two bodies whose masses are in ratio 2:1 are dropped simultaneoutwo places A and B where the acceleration due to gravity is gA and gB .if they reach ground at same time the ratio of heights from which they are dropped

Answer 1

Given:

Two bodies whose masses are in ratio 2:1 are dropped simultaneously at two places A and B where the acceleration due to gravity is gA and gB . They reach the ground at same time.

To find:

Ratio of heights from which they were released.

Calculation:

Since the acceleration at both the places are considered to be constant, we can easily apply equation of kinematics to solve this problem.

At place A :

$\therefore \: H_{A} = ut + \dfrac{1}{2} (g_{A} ){t}^{2}$

$= > \: H_{A} = 0+ \dfrac{1}{2} (g_{A} ){t}^{2}$

$= > \: H_{A} = \dfrac{1}{2} (g_{A}) {t}^{2}$

At place B :

$\therefore \: H_{B} = ut + \dfrac{1}{2} (g_{B}) {t}^{2}$

$= > \: H_{B} = 0 + \dfrac{1}{2} (g_{B}) {t}^{2}$

$= > \: H_{B} = \dfrac{1}{2} (g_{B}) {t}^{2}$

Hence the required ratio is :

$= > H_{A} : H_{B} = \dfrac{1}{2} (g_{A} ){t}^{2} : \dfrac{1}{2} (g_{B}) {t}^{2}$

$= > H_{A} : H_{B} = g_{A} : g_{B}$

So final answer :

$\boxed{ \bold{ H_{A} : H_{B} = g_{A} : g_{B}}}$

Answer 2

Answer:

gA:gB

Explanation:

we know that for vertical motion u=0 and

case:1

hA=Ut+1/2ga t^2

hA=1/2gA t^2

case:2

hB=ut+1/2gB t^2

hB=1/2gb t^2

so hA/hB=ga/gb