Question

If the time of fall of the two objects are in the ratio 1:2 to find the ratio of the height from which they fall
A1:2. B2:1. C1:4. D4:1

Answer 1

mass of 'A' is Ma
mass of 'B' is Mb
assume that
one of height is a and of another is b
both of them undergo an accelaration of 'g' as it is falling from a height
initial velocity of both is zero as it is dropped from rest.
From 2nd equation of motion
S = 1/2 a t2                      [ since u = 0 , it is not taken. ]
a = 1/2g (t1)^2
b = 1/2g (t2)^2
a/b = (t1 / t2)^2a/b = (1/2)^2  = 1/4he ratio of the height from which they fall = 1:4

i hope it will help you
regards

Answer 2

Answer:

So required ratio of height is 1:4

Explanation:

Let height of first object be x and height of second object be y.

Both of them undergo an acceleration of 'g' as it is falling from a height initial velocity of both object is zero because both are drop from rest.

By second equation of motion,

$S = u + \frac{1}{2} a {t}^{2}$

$S = \frac{1}{2} a {t}^{2}$

Here u=0

Now,

$x= \frac{1}{2} g {t1}^{2}$

and

$y = \frac{1}{2} g {t2}^{2}$

$\frac{x}{y} = {(\frac{t1}{t2})}^{2} = { \frac{1}{2} }^{2} = \frac{1}{4}$

So required ratio of height is 1:4.